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# 1.7: Trigonometric Functions of Any Angle

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Identify the reference angles for angles in the unit circle.
• Identify the ordered pair on the unit circle for angles whose reference angle is \begin{align*}30^\circ\end{align*}, \begin{align*}45^\circ\end{align*}, and \begin{align*}60^\circ\end{align*}, or a quadrantal angle, including negative angles, and angles whose measure is greater than \begin{align*}360^\circ\end{align*}.
• Use these ordered pairs to determine values of trig functions of these angles.
• Use calculators to find values of trig functions of any angle.

## Reference Angles and Angles in the Unit Circle

In the previous lesson, one of the review questions asked you to consider the angle \begin{align*}150^\circ\end{align*}. If we graph this angle in standard position, we see that the terminal side of this angle is a reflection of the terminal side of \begin{align*}30^\circ\end{align*}, across the \begin{align*}y-\end{align*}axis.

Notice that \begin{align*}150^\circ\end{align*} makes a \begin{align*}30^\circ\end{align*} angle with the negative \begin{align*}x-\end{align*}axis. Therefore we say that \begin{align*}30^\circ\end{align*} is the reference angle for \begin{align*}150^\circ\end{align*}. Formally, the reference angle of an angle in standard position is the angle formed with the closest portion of the \begin{align*}x-\end{align*}axis. Notice that \begin{align*}30^\circ\end{align*} is the reference angle for many angles. For example, it is the reference angle for \begin{align*}210^\circ\end{align*} and for \begin{align*}-30^\circ\end{align*}.

In general, identifying the reference angle for an angle will help you determine the values of the trig functions of the angle.

Example 1: Graph each angle and identify its reference angle.

a. \begin{align*}140^\circ\end{align*}

b. \begin{align*}240^\circ\end{align*}

c. \begin{align*}380^\circ\end{align*}

Solution:

a. \begin{align*}140^\circ\end{align*} makes a \begin{align*}40^\circ\end{align*} angle with the \begin{align*}x-\end{align*}axis. Therefore the reference angle is \begin{align*}40^\circ\end{align*}.

b. \begin{align*}240^\circ\end{align*} makes a \begin{align*}60^\circ\end{align*} with the \begin{align*}x-\end{align*}axis. Therefore the reference angle is \begin{align*}60^\circ\end{align*}.

c. \begin{align*}380^\circ\end{align*} is a full rotation of \begin{align*}360^\circ\end{align*}, plus an additional \begin{align*}20^\circ\end{align*}. So this angle is co-terminal with \begin{align*}20^\circ\end{align*}, and \begin{align*}20^\circ\end{align*} is its reference angle.

If an angle has a reference angle of \begin{align*}30^\circ\end{align*}, \begin{align*}45^\circ\end{align*}, or \begin{align*}60^\circ\end{align*}, we can identify its ordered pair on the unit circle, and so we can find the values of the six trig functions of that angle. For example, above we stated that \begin{align*}150^\circ\end{align*} has a reference angle of \begin{align*}30^\circ\end{align*}. Because of its relationship to \begin{align*}30^\circ\end{align*}, the ordered pair for is \begin{align*}150^\circ\end{align*} is \begin{align*}\left ( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right )\end{align*}. Now we can find the values of the six trig functions of \begin{align*}150^\circ\end{align*}:

\begin{align*}\cos 150 & = x = \frac{-\sqrt{3}}{2} && \sec 150 = \frac{1}{x} = \frac{1}{\frac{-\sqrt{3}}{2}} = \frac{-2}{\sqrt{3}}\\ \sin 150 & = y = \frac{1}{2} && \csc 150 = \frac{1}{y} = \frac{1}{\frac{1}{2}} = 2\\ \tan 150 & = \frac{y}{x} = \frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}} = \frac{1}{-\sqrt{3}} && \cot 150 = \frac{x}{y} = \frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3}\end{align*}

Example 2: Find the ordered pair for \begin{align*}240^\circ\end{align*} and use it to find the value of sin \begin{align*}240^\circ\end{align*}.

Solution: \begin{align*}\sin 240^\circ = \frac{-\sqrt{3}}{2}\end{align*}

As we found in example 1, the reference angle for \begin{align*}240^\circ\end{align*} is \begin{align*}60^\circ\end{align*}. The figure below shows \begin{align*}60^\circ\end{align*} and the three other angles in the unit circle that have \begin{align*}60^\circ\end{align*} as a reference angle.

The terminal side of the angle \begin{align*}240^\circ\end{align*} represents a reflection of the terminal side of \begin{align*}60^\circ\end{align*} over both axes. So the coordinates of the point are \begin{align*}\left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right )\end{align*}. The \begin{align*}y-\end{align*}coordinate is the sine value, so \begin{align*}\sin 240^\circ = \frac{-\sqrt{3}}{2}\end{align*}.

Just as the figure above shows \begin{align*}60^\circ\end{align*} and three related angles, we can make similar graphs for \begin{align*}30^\circ\end{align*} and \begin{align*}45^\circ\end{align*}.

Knowing these ordered pairs will help you find the value of any of the trig functions for these angles.

Example 3: Find the value of \begin{align*}\cot 300^\circ\end{align*}

Solution: \begin{align*}\cot 300^\circ = -\frac{1}{\sqrt{3}}\end{align*}

Using the graph above, you will find that the ordered pair is \begin{align*}\left ( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right )\end{align*}. Therefore the cotangent value is \begin{align*}\cot 300 = \frac{x}{y} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{2} \times -\frac{2}{\sqrt{3}} = -\frac{1}{\sqrt{3}}\end{align*}

We can also use the concept of a reference angle and the ordered pairs we have identified to determine the values of the trig functions for other angles.

## Trigonometric Functions of Negative Angles

Recall that graphing a negative angle means rotating clockwise. The graph below shows \begin{align*}-30^\circ\end{align*}.

Notice that this angle is coterminal with \begin{align*}330^\circ\end{align*}. So the ordered pair is \begin{align*}\left ( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right )\end{align*}. We can use this ordered pair to find the values of any of the trig functions of \begin{align*}-30^\circ\end{align*}. For example, \begin{align*}\cos (-30^\circ) = x = \frac{\sqrt{3}}{2}\end{align*}.

In general, if a negative angle has a reference angle of \begin{align*}30^\circ\end{align*}, \begin{align*}45^\circ\end{align*}, or \begin{align*}60^\circ\end{align*}, or if it is a quadrantal angle, we can find its ordered pair, and so we can determine the values of any of the trig functions of the angle.

Example 4: Find the value of each expression.

a. \begin{align*}\sin(-45^\circ)\end{align*}

b. \begin{align*}\sec(-300^\circ)\end{align*}

c. \begin{align*}\cos(-90^\circ)\end{align*}

Solution:

a. \begin{align*}\sin (-45^\circ) = -\frac{\sqrt{2}}{2}\end{align*}

\begin{align*}-45^\circ\end{align*} is in the \begin{align*}4^{th}\end{align*} quadrant, and has a reference angle of \begin{align*}45^\circ\end{align*}. That is, this angle is coterminal with \begin{align*}315^\circ\end{align*}. Therefore the ordered pair is \begin{align*}\left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right )\end{align*} and the sine value is \begin{align*}-\frac{\sqrt{2}}{2}\end{align*}.

b. \begin{align*}\sec(-300^\circ) = 2\end{align*}

The angle \begin{align*}-300^\circ\end{align*} is in the \begin{align*}1^{st}\end{align*} quadrant and has a reference angle of \begin{align*}60^\circ\end{align*}. That is, this angle is coterminal with \begin{align*}60^\circ\end{align*}. Therefore the ordered pair is \begin{align*}\left ( \frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*} and the secant value is \begin{align*}\frac{1}{x} = \frac{1}{\frac{1}{2}} = 2\end{align*}.

c. \begin{align*}\cos(-90^\circ) = 0\end{align*}

The angle \begin{align*}-90^\circ\end{align*} is coterminal with \begin{align*}270^\circ\end{align*}. Therefore the ordered pair is (0, -1) and the cosine value is 0.

We can also use our knowledge of reference angles and ordered pairs to find the values of trig functions of angles with measure greater than 360 degrees.

## Trigonometric Functions of Angles Greater than 360 Degrees

Consider the angle \begin{align*}390^\circ\end{align*}. As you learned previously, you can think of this angle as a full 360 degree rotation, plus an additional 30 degrees. Therefore \begin{align*}390^\circ\end{align*} is coterminal with \begin{align*}30^\circ\end{align*}. As you saw above with negative angles, this means that \begin{align*}390^\circ\end{align*} has the same ordered pair as \begin{align*}30^\circ\end{align*}, and so it has the same trig values. For example,

\begin{align*}\cos 390^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}\end{align*}

In general, if an angle whose measure is greater than \begin{align*}360\end{align*} has a reference angle of \begin{align*}30^\circ\end{align*}, \begin{align*}45^\circ\end{align*}, or \begin{align*}60^\circ\end{align*}, or if it is a quadrantal angle, we can find its ordered pair, and so we can find the values of any of the trig functions of the angle. Again, determine the reference angle first.

Example 5: Find the value of each expression.

a. \begin{align*}\sin 420^\circ\end{align*}

b. \begin{align*}\tan 840^\circ\end{align*}

c. \begin{align*}\cos 540^\circ\end{align*}

Solution:

a. \begin{align*}\sin 420^\circ = \frac{\sqrt{3}}{2}\end{align*}

\begin{align*}420^\circ\end{align*} is a full rotation of 360 degrees, plus an additional 60 degrees. Therefore the angle is coterminal with \begin{align*}60^\circ\end{align*}, and so it shares the same ordered pair, \begin{align*}\left ( \frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*}. The sine value is the \begin{align*}y-\end{align*}coordinate.

b. \begin{align*}\tan 840^\circ = -\sqrt{3}\end{align*}

\begin{align*}840^\circ\end{align*} is two full rotations, or 720 degrees, plus an additional 120 degrees:

\begin{align*}840 = 360 + 360 + 120\end{align*}

Therefore \begin{align*}840^\circ\end{align*} is coterminal with \begin{align*}120^\circ\end{align*}, so the ordered pair is \begin{align*}\left ( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*}. The tangent value can be found by the following:

\begin{align*}\tan 840^\circ = \tan 120^\circ = \frac{y}{x} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{2} \times -\frac{2}{1} = -\sqrt{3}\end{align*}

c. \begin{align*}\cos 540^\circ = -1\end{align*}

\begin{align*}540^\circ\end{align*} is a full rotation of 360 degrees, plus an additional 180 degrees. Therefore the angle is coterminal with \begin{align*}180^\circ\end{align*}, and the ordered pair is (-1, 0). So the cosine value is -1.

So far all of the angles we have worked with are multiples of 30, 45, 60, and 90. Next we will find approximate values of the trig functions of other angles.

## Using a Calculator to Find Values

If you have a scientific calculator, you can determine the value of any trig function for any angle. Here we will focus on using a TI graphing calculator to find values.

First, your calculator needs to be in the correct “mode.” In chapter 2 you will learn about a different system for measuring angles, known as radian measure. In this chapter, we are measuring angles in degrees. We need to make sure that the calculator is in degrees. To do this, press \begin{align*}\fbox{\text{MODE}}\end{align*}. In the third row, make sure that Degree is highlighted. If Radian is highlighted, scroll down to this row, scroll over to Degree, and press \begin{align*}\fbox{\text{ENTER}}\end{align*}. This will highlight Degree. Then press \begin{align*} \begin{array} {|c|} \hline 2^{\text{nd}} \\ \hline \end{array}\end{align*} \begin{align*}\fbox{\text{MODE}}\end{align*} to return to the main screen.

Now you can calculate any value. For example, we can verify the values from the table above. To find \begin{align*}\sin 130^\circ\end{align*}, press \begin{align*}\fbox{\text{Sin}}\end{align*} \begin{align*}\fbox{130}\end{align*} \begin{align*}\fbox{\text{ENTER}}\end{align*}. The calculator should return the value .7660444431.

Example 6: Find the approximate value of each expression. Round your answer to 4 decimal places.

a. \begin{align*}\sin 130^\circ\end{align*}

b. \begin{align*}\cos 15^\circ\end{align*}

c. \begin{align*}\tan 50^\circ\end{align*}

Solution:

a. \begin{align*}\sin 130^\circ \approx 0.7660\end{align*}

b. \begin{align*}\cos 15^\circ \approx 0.9659\end{align*}

c. \begin{align*}\tan 50^\circ \approx 1.1918\end{align*}

You may have noticed that the calculator provides a “(“ after the SIN. In the previous calculations, you can actually leave off the “)”. However, in more complicated calculations, leaving off the closing “)” can create problems. It is a good idea to get in the habit of closing parentheses.

You can also use a calculator to find values of more complicated expressions.

Example 7: Use a calculator to find an approximate value of \begin{align*}\sin 25^\circ + \cos 25^\circ\end{align*}. Round your answer to 4 decimal places.

Solution: \begin{align*}\sin 25^\circ + \cos 25^\circ \approx 1.3289\end{align*}

\begin{align*}^* \text{This is an example where you need to close the parentheses.}\end{align*}

## Points to Consider

• What is the difference between the measure of an angle, and its reference angle? In what cases are these measures the same value?
• Which angles have the same cosine value, or the same sine value? Which angles have opposite cosine and sine values?

## Review Questions

1. State the reference angle for each angle.
1. \begin{align*}190^\circ\end{align*}
2. \begin{align*}-60^\circ\end{align*}
3. \begin{align*}1470^\circ\end{align*}
4. \begin{align*}-135^\circ\end{align*}
2. State the ordered pair for each angle.
1. \begin{align*}300^\circ\end{align*}
2. \begin{align*}-150^\circ\end{align*}
3. \begin{align*}405^\circ\end{align*}
3. Find the value of each expression.
1. \begin{align*}\sin 210^\circ\end{align*}
2. \begin{align*}\tan 270^\circ\end{align*}
3. \begin{align*}\csc 120^\circ\end{align*}
4. Find the value of each expression.
1. \begin{align*}\sin 510^\circ\end{align*}
2. \begin{align*}\cos 930^\circ\end{align*}
3. \begin{align*}\csc 405^\circ\end{align*}
5. Find the value of each expression.
1. \begin{align*}\cos -150^\circ\end{align*}
2. \begin{align*}\tan -45^\circ\end{align*}
3. \begin{align*}\sin -240^\circ\end{align*}
6. Use a calculator to find each value. Round to 4 decimal places.
1. \begin{align*}\sin 118^\circ\end{align*}
2. \begin{align*}\tan 55^\circ\end{align*}
3. \begin{align*}\cos 100^\circ\end{align*}
7. Recall, in lesson 1.4, we introduced inverse trig functions. Use your calculator to find the measure of an angle whose sine value is 0.2.
8. In example 6c, we found that \begin{align*}\tan 50^\circ \approx 1.1918\end{align*}. Use your knowledge of a special angle to explain why this value is reasonable. HINT: You will need to change the tangent of this angle into a decimal.
9. Use the table below or a calculator to explore sum and product relationships among trig functions. Consider the following functions: \begin{align*}f(x) & = \sin (x+x) \ \text{and} \ g(x) = \sin (x) + \sin (x)\\ h(x) & = \sin (x) \ ^* \ \sin(x) \ \text{and} \ j(x) = \sin (x^2)\end{align*} Do you observe any patterns in these functions? Are there any equalities among the functions? Can you make a general conjecture about \begin{align*}\sin (a) + \sin (b)\end{align*} and \begin{align*}\sin (a+b)\end{align*} for all values of \begin{align*}a, b\end{align*}? What about \begin{align*}\sin(a) \sin(a)\end{align*} and \begin{align*}\sin (a^2)\end{align*}?
\begin{align*}a^\circ\end{align*} \begin{align*}b^\circ\end{align*} \begin{align*}\sin a + \sin b\end{align*} \begin{align*}\sin(a + b)\end{align*}
10 30 .6736 .6428
20 60 1.2080 .9848
55 78 1.7973 .7314
122 25 1.2707 .5446
200 75 .6239 -.9962
1. Use a calculator or your knowledge of special angles to fill in the values in the table, then use the values to make a conjecture about the relationship between \begin{align*}(\sin a)^2\end{align*} and \begin{align*}(\cos a)^2\end{align*}. If you use a calculator, round all values to 4 decimal places.
\begin{align*}a\end{align*} \begin{align*}(\sin a)^2\end{align*} \begin{align*}(\cos a)^2\end{align*}
0
25
45
80
90
120
250

1. \begin{align*}10^\circ\end{align*}
2. \begin{align*}60^\circ\end{align*}
3. \begin{align*}30^\circ\end{align*}
4. \begin{align*}45^\circ\end{align*}
1. \begin{align*}\left ( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right )\end{align*}
2. \begin{align*}\left ( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right )\end{align*}
3. \begin{align*}\left ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right )\end{align*}
1. \begin{align*}-\frac{1}{2}\end{align*}
2. 0
3. \begin{align*}\frac{2}{\sqrt{3}}\end{align*}
1. \begin{align*}\frac{1}{2}\end{align*}
2. \begin{align*}-\frac{\sqrt{3}}{2}\end{align*}
3. \begin{align*}\sqrt{2}\end{align*}
1. \begin{align*}-\frac{\sqrt{3}}{2}\end{align*}
2. -1
3. \begin{align*}\frac{\sqrt{3}}{2}\end{align*}
1. 0.8828
2. 1.4281
3. -0.1736
2. This is reasonable because \begin{align*}\tan 45^\circ = 1\end{align*} and the \begin{align*}\tan 60^\circ = \sqrt{3} \approx 1.732\end{align*}, and the \begin{align*}\tan 50^\circ\end{align*} should fall between these two values.
3. Conjecture: \begin{align*}\sin a + \sin b \ne \sin(a + b)\end{align*}
\begin{align*}a\end{align*} \begin{align*}(\sin a)^2\end{align*} \begin{align*}(\cos a)^2\end{align*}
0 0 1
25 .1786 .8216
45 \begin{align*}\frac{1}{2}\end{align*} \begin{align*}\frac{1}{2}\end{align*}
80 .9698 .0302
90 1 0
120 .75 .25
235 .6710 .3290
310 .5898 .4132

Conjecture: \begin{align*}(\sin a)^2 + (\cos a)^2 = 1\end{align*}.

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# Trying to define $\mathbb{R}^{0.5}$ topologically [duplicate]

A few days ago, I was trying to generalize the defintion of Euclidean spaces by trying to define $\mathbb{R}^{0.5}$.

Question: Is there a metric space $A$ such that $A\times A$ is homeomorphic to $\mathbb{R}$?

I am interested also in seeing examples of $A$ which are only topological spaces

Edit: If there exists a topological space $A$ such that $A\times A\cong \Bbb R$, then $A\times \{a\}$ is a subspace of $A\times A$ ($a\in A$). Hence $A\times\{a\}$ can be embedded in $\mathbb{R}$, since $A\cong A\times \{a\}$. Thus $A$ can be embedded in $\mathbb{R}$. Therefore $A$ is metrizable.

Thank you

• do you want $A$ to be a metric space or a topological space? – Ittay Weiss Mar 12 '13 at 9:21
• @IttayWeiss I asked for a metric space, but I would be very intrested to see such a topological space – Amr Mar 12 '13 at 9:22
• @IttayWeiss $A$ has to be a metric space. Please read my question again to see my argumet. – Amr Mar 12 '13 at 9:34
• Nice to see a heavily upvoted question about a more advanced topic for a change (even if it's one I don't completely understand). – Joe Z. Mar 12 '13 at 19:32
• Duplicate of math.stackexchange.com/q/57375 – Martin Mar 29 '13 at 9:36

## 3 Answers

No, no such space exists. Suppose $A$ is a topological space such that $A\times A\cong \mathbb R$, with the Euclidean induced topology on $\mathbb R$. If $A$ is disconnected then so is $A\times A$, but that would contradict the connectivity of $\mathbb R$, so it follows that $A$ is connected. But, since $A$ is connected it follows that $A\times A$ with a single point removed is still connected. However, $\mathbb R$ with a single point removed is not connected. Contradiction.

• A very easy yet fascinating argument! +1 upvote, of course. – Sangchul Lee Mar 12 '13 at 9:48
• $A$ is the continuous image of $\mathbb{R}$ under projection so must be connected. This simplifies the argument a little. – Henno Brandsma Mar 12 '13 at 14:07
• Hi Ittay. Maybe this question is just triviality, but could you clarify why $A\times A$ with a single point removed is still connected? I know if $A=\mathbb{R}$ then one can either produce arguments with topological dimensions or use path connectedness. But for any connected space $A$ I can't see how it follows immediately. – T. Eskin Mar 12 '13 at 18:25
• @ThomasE. you can express it as a union of connected parts with connected intersections. Alternatively, the whole argument can be repeated with path connectedness instead of connectedness. Henno's argument also shows that $A$ must be path connected, since it's a continuous image of $\mathbb R$. – Ittay Weiss Mar 12 '13 at 18:37

If $A \times A \simeq \mathbb{R}$, then $A$ has to be connected and it is homeomorphic to a subspace of $\mathbb{R}$, so $A$ is homeomorphic to an interval. Therefore, either $A \times A \simeq [0,1]^2$ or $A \times A \simeq (0,1)^2 \simeq \mathbb{R}^2$ or $A \times A \simeq [0,1)^2$. Thus, $A \times A$ can't be homeomorphic to $\mathbb{R}$.

• You should probably say some things about why those spaces are not homeomorphic to $\mathbb{R}$. The easiest argument is probably that none of them can be disconnected by removing a point, which is just Ittay Weiss' argument. – Qiaochu Yuan Mar 13 '13 at 2:09

This was intended as a comment on the answers above, but I'm a new user, so I don't have that as an option, so I'll need to post it as a new answer instead.

The other answers here show that $\mathbb{R}$ cannot be written as a "square" of sorts. The situation is not unlike that of the real number -1, which cannot be written as a square of real numbers.

But one can extend the real numbers to create a square root of $-1$. Similarly, one could extend the notion of topological space to create a "square root" of $\mathbb{R}$. One way to do so, in analogy with the complex numbers, would be as follows. Define a "complex topological space" to be an ordered pair $(A,B)$ of topological spaces. Define the direct product of two such ordered pairs by

$(A,B) \times (C,D)$ equals (disjoint union of $A\times C$ and $\mathbb{R}\times B \times D$, disjoint union of $A\times D$ and $B\times C$)

Identify a topological space $Y$ with the ordered pair $(Y, \varnothing)$.

Then $(\varnothing$, one-point set$)\times (\varnothing$, one-point set) = ($\mathbb{R}$, empty set).

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# Lecture 6: Monty Hall, Simpson's Paradox

## The Monty Hall Problem

In case you did not grow up watching way too much daytime television in America during the 70's and early 80's, here is Monty Hall on YouTube talking about the background of this math problem involving his popular game show, Let's Make A Deal.

• There are three doors.
• A car is behind one of the doors.
• The other two doors have goats behind them.
• You choose a door, but before you see what's behind your choice, Monty opens one of the other doors to reveal a goat.
• Monty offers you the chance to switch doors.

Should you switch?

### Defining the problem

Let $S$ be the event of winning when you switch.

Let $D_j$ be the event of the car being behind door $j$.

### Solving with a probability tree

With a probability tree, it is easy to represent the case where you condition on Monty opening door 2. Given that you initially choose door 1, you can quickly see that if you stick with door 1, you have a $\frac{1}{3}~$ chance of winning.

You have a $\frac{2}{3}~$ chance of winning if you switch.

### Solving with the Law of Total Probability

This is even easier to solve using the Law of Total Probability.

\begin{align} P(S) &= P(S|D_1)P(D_1) + P(S|D_2)P(D_2) + P(S|D_3)P(D_3) \\ &= 0 \frac{1}{3} + 1 \frac{1}{3} + 1 \frac{1}{3} \\ &= \frac{2}{3} \end{align}

### A more general solution

Let $n = 7$ be the number of doors in the game.

Let $m=3$ be the number of doors with goats that Monty opens after you select your initial door choice.

Let $S$ be the event where you win by sticking with your original door choice of door 1.

Let $C_j$ be the event that the car is actually behind door $j$.

Conditioning only on which door has the car, we have \begin{align} & &P(S) &= P(S|C_1)P(C_1) + \dots + P(S|C_n)P(C_n) & &\text{Law of Total Probability} \\ & & &= P(C_1) \\ & & &= \frac{1}{7} \\ \end{align}

Let $M_{i,j,k}$ be the event that Monty opens doors $i,j,k$. Conditioning on Monty opening up doors $i,j,k$, we have

\begin{align} & &P(S) &= \sum_{i,j,k} P(S|M_{i,j,k})P(M_{i,j,k}) & &\text{summed over all i, j, k with } 2 \le i \lt j \lt k \le 7 \\ \\ & &\Rightarrow P(S|M_{i,j,k}) &= P(S) & &\text{by symmetry} \\ & & &=\frac{1}{7} \end{align}

Note that we can now generalize this to the case where:

• there are $n \ge 3$ doors
• after you choose a door, Monty opens $m$ of the remaining doors $n-1$ doors to reveal a goat (with $1 \le m \le n-m-2$)

The probability of winning with the strategy of sticking to your initial choice is $\frac{1}{n}$, whether unconditional or conditioning on the doors Monty opens.

After Monty opens $m$ doors, each of the remaining $n-m-1$ doors has conditional probability of $\left(\frac{n-1}{n-m-1}\right) \left(\frac{1}{n}\right)$.

Since $\frac{1}{n} \lt \left(\frac{n-1}{n-m-1}\right) \left(\frac{1}{n}\right)$, you will always have a greater chance of winning if you switch.

Is it possible for a certain set of events to be more (or less) probable than another without conditioning, and then be less (or more) probable with conditioning?

Assume that we have the above rates of success/failure for Drs. Hibbert and Nick for two types of surgery: heart surgery and band-aid removal.

### Defining the problem

Let $A$ be the event of a successful operation.

Let $B$ be the event of treatment by Dr. Nick.

Let $C$ be the event of heart surgery.

\begin{align} P(A|B,C) &< P(A|B^c,C) & &\text{Dr. Nick is not as skilled as Dr. Hibbert in heart surgery} \\ P(A|B,C^c) &< P(A|B^c,C^c) & &\text{neither is he all that good at band-aid removal} \\ \end{align}

And yet $P(A|B) > P(A|B^c)$?

### Explaining with the Law of Total Probability

To explain this paradox, let's try to use the Law of Total Probability.

\begin{align} P(A|B) &= P(A|B,C)P(C|B) + P(A|B,C^c)P(C^c|B) \\ \\ \text{but } P(A|B,C) &< P(A|B^c,C) \\ \text{and } P(A|B,C^c) &< P(A|B^c,C^c) \end{align}

Look at $P(C|B$ and $P(C|B^c)$. These weights are what makes this paradox possible, as they are what make the inequality relation sign flip.

Event $C$ is a case of confounding

### Another example

_Is it possible to have events $A_1, A_2, B, C$ such that_

\begin{align} P(A_1|B) &\gt P(A_1|C) \text{ and } P(A_2|B) \gt P(A_2|C) & &\text{ ... yet...} \\ P(A_1 \cup A_2|B) &\lt P(A_1 \cup A_2|C) \end{align}

Yes, and this is just another case of Simpson's Paradox.

Note that

\begin{align} P(A_1 \cup A_2|B) &= P(A_1|B) + P(A_2|B) - P(A_1 \cap A_2|B) \end{align}

So this is not possible if $A_1$ and $A_2$ are disjoint and $P(A_1 \cup A_2|B) = P(A_1|B) + P(A_2|B)$.

It is crucial, therefore, to consider the intersection $P(A_1 \cap A_2|B)$, so let's look at the following example where $P(A_1 \cap A_2|B) \gg P(A_1 \cap A_2|C)$ in order to offset the other inequalities.

Consider two basketball players each shooting a pair of free throws.

Let $A_j$ be the event basketball free throw scores on the $j^{th}$ try.

Player $B$ always either makes both $P(A_1 \cap A_2|B) = 0.8$, or misses both.

\begin{align} P(A_1|B) = P(A_2|B) = P(A_1 \cap A_2|B) = P(A_1 \cup A_2|B) = 0.8 \end{align}

Player $C$ makes free throw shots with probability $P(A_j|C) = 0.7$, independently, so we have

\begin{align} P(A_1|C) &= P(A_2|C) = 0.7 \\ P(A_1 \cap A_2|C) &= P(A_1|C) P(A_2|C) = 0.49 \\ P(A_1 \cup A_2|C) &= P(A_1|C) + P(A_2|C) - P(A_1 \cap A_2|C) \\ &= 2 \times 0.7 - 0.49 \\ &= 0.91 \end{align}

And so we have our case where

\begin{align} P(A_1|B) = 0.8 &\gt P(A_1|C) = 0.7 \\ P(A_2|B) = 0.8 &\gt P(A_2|C) = 0.7 \\ \\ \text{ ... and yet... } \\ \\ P(A_1 \cup A_2|B) &\lt P(A_1 \cup A_2|C) ~~~~ \blacksquare \end{align}


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# Coordinating Computation

The general computational model underlying Zef is quite lazy. Much lazier than what you're used to from normal Python. It may seem a bit awkward and complicated at first, but it allows you to write very expressive and declarative code which is decoupled from the computation. Only at the very end if some output or action is requested, is any real work done. In general, only the minimal amount of work is done and this is determined by the Zef runtime, allowing you to focus on the logic when writing code to solve your problem.

Let's explore the different ways to trigger computation.
There are three general categories of

## Collect

• pure computation: only a value is wanted
• no other interactions with the outside world (aka side effects: e.g. sending an email, reading from disk, writing to a DB)
x = [2,3,4,5] | map[power[2]] | sum     # x is a LazyValue    # trigger compute: what is this concretely?  x | collect    # alternatives  collect(x)  x()

Note: in traditional Python a lazy value can be expressed by a pure, reproducible function which takes no arguments. As seen in the last line, LazyValues in Zef also drop out naturally as such: function that take no args.

## Executing Effects

Writing as much of your code as pure functions is one of the few things which most experienced developers can agree on as being a good thing. This applies especially to the parts containing the core logic of your application.
This is a noble goal, but in many cases the question is "what does this concretely mean?". "How can I do this principle to this very code I'm writing here?" ZefFX is a tool which can help you structure your application in this way by providing a friendly and convenient way to deal with side effects: a managed effect system.

my_wish = FX.HTTP.StartServer(port=8000)    # pure data    my_wish | run

You're handing your wish over to the runtime. Think of it as expressing the entire dirty task of interacting with the nasty outside world as data (sending it a message) and asking it "could you please do that for me?".
The runtime will respond with a message or an error. In case of an error, this should be descriptive.

### Executing Effect Streams

Suppose you have a Stream of wishes (effects). Each time a new wish flows through the stream, you want it to be executed by the runtime.
In this case you can simply "connect the hose to the runtime directly"

my_stream_of_wishes | subscribe

In contrast to run which takes a single or a fixed list of wishes and executes the entire list, subscribe acts asynchronously: each time a wish appears it gets executed directly, one after the other.

How to specify more advanced execution policies like

• handling back pressure
• handling wishes concurrently
will be discussed in more detail later.

Note: subscribe in it's raw form can be understood to have execute (the impure procedure that takes a wish and executes it) curried in as its default. You can also run you own impure procedures

my_stream | subscribe[print]

will hook up the stream, activate the pipeline and call the (impure) function print on each value as it appears in my_stream.


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## Encyclopedia > Conjugation (grammar)

Article Content

# Grammatical conjugation

Redirected from Conjugation (grammar)

Grammatical conjugation is the creation of derived forms of a verb from one basic form. Conjugation may be affected by person, number, gender, tense, mood, voice, or some other language-specific factor or factors. When a verb is used to function as the action done by a subject, the verb must be conjugated in most languages.

For example, here's a sample conjugation of the English verb to be and its Latin and French equivalents, esse and être. Except for the infinitive, which is given in the present form, all the verbs listed are in the present indicative active.

to be / esse / être
Form / Person English Latin French
Infinitive to be esse être
1st singular I am sum je suis
2nd singular you are es tu es
3rd singular he, she, or it is est il/elle est
1st plural we are sumus nous sommes
2nd plural you are estis vous êtes
3rd plural they are sunt ils/elles sont

All Wikipedia text is available under the terms of the GNU Free Documentation License

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In competitive swimming, the medley relay is a team event in which four different swimmers each swim one of the four strokes: freestyle, breaststroke, backstroke, and butterfly. In general, every swimmer might be able swim any given stroke. How can we choose the fastest relay team? Here we solve this by enumerating all possible teams, though a more efficient routine likely exists.

Some example practice times can be seen on this Google Spreadsheet. Also, using the steps outlined here, the same spreadsheet is available as a CSV file here. (FTR, these are actual practice times for 100 yards for mostly 12-13 year-old swimmers; the names have been changed.)

SAS
We first read the data from the URL, using the technique outlined in section 1.1.6. Note that if you cut-and-paste this, you’ll need to get the whole URL onto one line– we break it up here for display only.
filename swimurl url 'https://docs.google.com/spreadsheet
/pub?key=0AvJKgZUzMYLYdE5xTHlEWkNUM3NoOHB1ZTJoTFMzUUE&
single=true&gid=0&output=csv';

proc import datafile=swimurl out=swim dbms=csv;
run;

Next, we use the point= option in nested set statements to generate a single data set with all the possible combinations of names and times. Meanwhile we change the names of the variables so they don’t get overwritten in the next set statement. Note the use of the nobs option to find the number of rows in the data set.
data permute
(keep=free freetime fly flytime back backtime  breast breasttime);
set swim (rename = (swimmer=free freestyle=freetime)) nobs=nobs;
do i = 1 to nobs;
set swim(rename = (swimmer=fly butterfly=flytime)) point=i;
do j = 1 to nobs;
set swim(rename = (swimmer=back backstroke=backtime)) point=j;
do k = 1 to nobs;
set swim(rename = (swimmer=breast breaststroke=breasttime))
point = k;
output;
end;
end;
end;
run;

The resulting data set has 12^4 rows, and includes rosters with the same swimmer swimming all four legs. In fact, a quick glance will show that Anna has the best time in each stroke, and thus the best “team” based on these practice times would use her for each stroke. This is against the rules, and also probably isn’t reflective of performance in a race. We’ll remove illegal line-ups using a where statement (section 1.5.1) and also calculate the total team time.
data prep;
set permute;
where free ne back and free ne breast and free ne fly and
back ne breast and back ne fly and breast ne fly;
time = sum(freetime, flytime, backtime, breasttime);
run;

The resulting data set has (12 permute 4) lines. To find the best team, we just sort by the total time and look at the first line. Here the first 10 lines (10 best teams) are shown.
proc sort data=prep; by time; run;
proc print data=prep (obs=10); run;

b
r
f                             b              e
r              f              a              a
e              l              c      b       s
e              y              k      r       t
f       t              t      b       t      e       t      t
r       i      f       i      a       i      a       i      i
e       m      l       m      c       m      s       m      m
e       e      y       e      k       e      t       e      e

Kara    109.3  Dora    126.8  Lara    117.7  Anna    126.9  480.7
Anna    102.8  Dora    126.8  Lara    117.7  Beth    134.6  481.9
Kara    109.3  Anna    120.5  Lara    117.7  Beth    134.6  482.1
Anna    102.8  Dora    126.8  Lara    117.7  Honora  136.4  483.7
Kara    109.3  Jane    129.8  Lara    117.7  Anna    126.9  483.7
Kara    109.3  Anna    120.5  Lara    117.7  Dora    136.4  483.9
Kara    109.3  Anna    120.5  Lara    117.7  Honora  136.4  483.9
Kara    109.3  Lara    123.1  Jane    124.7  Anna    126.9  484.0
Anna    102.8  Dora    126.8  Lara    117.7  Inez    136.8  484.1
Carrie  112.7  Dora    126.8  Lara    117.7  Anna    126.9  484.1

The best time shaves a whole second off the predicted time using the second-best team.

R
Since published Google Spreadsheets use https rather than http, we use the RCurl package and its getURL() function. (Note that if you cut-and-paste this, you’ll need to get the whole URL onto one line– we break it up here for display only.) Then we can read the data with read.csv() and textConnection().
library(RCurl)
/pub?key=0AvJKgZUzMYLYdE5xTHlEWkNUM3NoOHB1ZTJoTFMzUUE&
single=true&gid=0&output=csv")

To make an object with the combinations of names, we use the expand.grid() function highlighted in Example 7.22, providing as arguments the swimmers names four times. As in the SAS example, the result has has 12^4 rows. The combn() function might be a better fit here, but was more difficult to use.
test2 = expand.grid(swim2$Swimmer,swim2$Swimmer, swim2$Swimmer, swim2$Swimmer)

It’ll be useful to assign these copies of the names to each of the strokes. We’ll do that with the rename() function available in the reshape package. (This approach is mentioned in section 1.3.4.). Then we can remove the rows where the same name appears twice using some logic. The logic is nested in the with() function to save some keystrokes and is generally preferable to attach()ing the test2 object.
library(reshape)
test2 = rename(test2, c("Var1" = "free", "Var2" = "fly",
"Var3" = "back", "Var4" = "breast"))
test3 = with(test2, test2[(free != breast) & (free != fly)
& (free != back) & (breast != fly) & (breast != back)
& (fly != back) ,])

Finally, we can use the which.min() function to pick the best team.
> bestteam =
+ test3[which.min(swim2$Freestyle[test3$free]+swim2$Breaststroke[test3$breast] +
+ swim2$Butterfly[test3$fly] + swim2$Backstroke[test3$back]),]
>bestteam
free  fly back breast
1631 Kara Dora Lara   Anna

For new users of R, this may look very peculiar– it uses elegant but powerful features of the R language that may be challenging for new users to grasp. Essentially, in swim2$Freestyle[test3$free] we say: from the “freestyle” times in the swim2 object, take the time from the row that has the name in a row of “free” names in the test3 object. The which.min() function replicates this request for every row in the test3 object (which has all of the permutations) in it, returning the row number with that minimum sum. The outer test3[rows,columns] syntax grabs the values in this row. (The number 1631 is the row number, for some reason showing the row in the test2 object created by expand.grid().)

Now, we might also want the actual times associated with the best team. We can find them by calling the correct rows (names from the best team) and columns (stroke associated with that name) from the original data set.
> times = c(swim2[swim2$Swimmer == bestteam$free,2],
+      swim2[swim2$Swimmer == bestteam$fly,3],
+      swim2[swim2$Swimmer == bestteam$back,4],
+      swim2[swim2$Swimmer == bestteam$breast,5])
> times
[1] 109.3 126.8 117.7 126.9

If instead, one wanted to list the times in order, one approach would be to add columns to the test3 object with the time for each stroke, calculate their sum, and sort on the sum.


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# Into Math Grade 4 Module 13 Lesson 4 Answer Key Relate Degrees to Fractional Parts of a Circle

We included HMH Into Math Grade 4 Answer Key PDF Module 13 Lesson 4 Relate Degrees to Fractional Parts of a Circle to make students experts in learning maths.

## HMH Into Math Grade 4 Module 13 Lesson 4 Answer Key Relate Degrees to Fractional Parts of a Circle

I Can find the measure of an angle by using its fractional part of a circle.

Based on what you know about the measurement of the shaded piece of the circle, how can you describe the measurement of the circle?

90 degrees
Explanation:
Angle of complete circle is 360 degrees.
4 equal right angle make a circle.
Angle of right angle is 90 degrees.
So, the shaded part is 90 degrees.

Turn and Talk What other fractions can you use to describe the measurement of a circle?
The measure of an angle is determined by the amount of rotation from the initial side to the terminal side.
One way to measure an angle is in terms of degrees .
A measure of one degree ( 1° ) is equivalent to a rotation of 1360 of a complete revolution.

Build Understanding

Question 1.
Circles can be divided into measurement units called degrees. How can you relate the fractional turns of a circle to degrees?
90 degrees
Explanation:
Angles are measured in units called degrees. The symbol for degrees is °.
If a circle is divided into 360 equal parts, then an angle that turns through 1 part of the 360 measures 1°
.
Fraction is a part of a whole.

Connect to Vocabulary
Angles are measured with reference to a circle as they turn around a center point. The size of the angle is the measure of how far one side is turned from the other side. Angles are measured in units called degrees. The symbol for degrees is °.

A. How can you write a fraction to represent the quarter turns in a circle?

• _______ is a quarter turn.
• _______ is two quarter turns.
• _______ is three quarter turns.
• _______ is four quarter turns.
_______ quarter turns make a full turn on a circle.

• 90° is a quarter turn.
• 180° is two quarter turns.
• 270° is three quarter turns.
• 360° is four quarter turns.
four quarter turns make a full turn on a circle.
Explanation:
Circle is divided into 4 quarter, each quarter is of 90 degrees.

B. How can you show the number of eighth turns it takes to make a full turn on a circle?
$$\frac{360}{8}$$ = 45°

Explanation:
360° full turn in the circle.
The number of eighth turns to make a full turn on a circle is $$\frac{360}{8}$$ = 45°

C. How many $$\frac{1}{120}$$ turns would it take to make a full turn around the circle?
$$\frac{1}{120}$$ turns

Explanation:
360° full turn in the circle.
The number of eighth turns to make a full turn on a circle is $$\frac{360}{3}$$ = 120°

D. How many $$\frac{1}{360}$$ turns would it take to make a full turn around the circle?
Think of a circle divided into 360 equal parts. An angle that turns through $$\frac{1}{360}$$ 0f the circle measures 1 degree.

360°
Explanation:
Angle of complete circle is 360 degrees, fraction of a whole is $$\frac{1}{360}$$
360° x $$\frac{1}{360}$$ = 1°

E. How can you use degrees to describe the measure of an angle that has eight $$\frac{1}{360}$$ turns?
45°
Explanation:
Angle of a circle is 360 degrees,
Eight turns of an angle is 45 degrees.
8 x $$\frac{1}{360}$$
= $$\frac{8}{360}$$
=45°

Question 2.
Drafters prepare technical drawings for all kinds of designs. A drafter draws this fractional part of a circle for a project. How can you find the measure of the angle in degrees?

A. How can you think about the fractional part of the circle as degrees?
40°
Explanation:
Angle of a circle is 360 degrees,
nine turns of an angle is 40 degrees
9 x $$\frac{360}{1}$$
= $$\frac{360}{9}$$
= 40°
B. How can you write $$\frac{1}{9}$$ as an equivalent fraction with a denominator of 360?
360 x $$\frac{1}{9}$$ = 40
Explanation:
Angle of a circle is 360 degrees,
nine turns of an angle is 40 degrees
9 x $$\frac{360}{1}$$
= $$\frac{360}{9}$$
= 40°

C. The measure of the angle is __________.
40 degrees
Explanation:
Angle of a circle is 360 degrees,
nine turns of an angle is 40 degrees
9 x $$\frac{360}{1}$$
= $$\frac{360}{9}$$
= 40°

Turn and Talk How can you describe the fraction of a circle through which a straight angle turns?
Angles are measured with reference to a circle as they turn around a center point.
The size of the angle is the measure of how far one side is turned from the other side.
Angles are measured in units called degrees.
The symbol for degrees is °.

Connect to Vocabulary
acute—an angle that measures greater than 0 and less than 90°
right—an angle that measures 90°
obtuse—an angle that measures greater than 90° and less than 180°
straight—an angle that measures 180°
reflex—an angle that measures greater than 180°

Question 3.
Angles are named by the way that they relate to 0°, 90°, or 180°. What is a name for each angle?

Explanation:
Angle less than 90 degrees is acute angle.
Angle greater than 90 degrees is obtuse angle.
Angle which measures 90 degrees is right angle.
Angle which measures 180 degrees is straight angle.
Angle which measures more than 180 degrees is reflex angle.

Turn and Talk How can you describe the fraction of a circle through which a straight angle turns?
A complete or whole circle is taken as 1 and parts of the circles are represented as fractions.
Explanation:
An angle is a fraction of a circle where the whole circle is 360°.
A straight angle is the same as half the circle and is 180° whereas a right angle is a quarter of a circle and is 90°. Use the protractor to measure the size of an angle.

Step It Out

Question 4.
Architects often use circles when designing buildings. How can you find the measure of an angle from the fractional part of the circle formed by the window?

A. To estimate, describe the angle.
30 degrees
Explanation:
Angle of complete circle is 360 degrees.
Whole is divided into 12 equal parts.
= $$\frac{360}{12}$$
= 30°

B. Determine the fraction of a circle that any one angle turns.
The fraction of a circle that any one angle turns

C. Write an equivalent fraction to find the measure of the angle in degrees.

Explanation:
Angle of complete circle is 360 degrees.
Whole is divided into 12 equal parts.
= $$\frac{360}{12}$$
= 30°

D. The measure of the angle from the fractional part of the circle formed by the window is _______.
30°
Explanation:
Angle of complete circle is 360 degrees.
Whole is divided into 12 equal parts.
= $$\frac{360}{12}$$
= 30°

Angles are measured with reference to a circle as they turn around a center point.
The size of the angle is the measure of how far one side is turned from the other side.
Angles are measured in units called degrees.
The symbol for degrees is °, here the measure of an angle is 30°

Check Understanding

Question 1.
Write the measure of the shaded angle in degrees.

Explanation:
The fraction of a circle that any one angle turns.
Complete angle of circle is 360 degrees.

Question 2.
Classify the angle. Write acute, right, obtuse, straight, or reflex.

Reflex angle
Explanation:
A reflex angle is any angle that is more than 180 degrees (half circle) and less than 360 degrees (full circle).
A reflex angle will always have either an obtuse or an acute angle on the other side of it.
It can be one of the more confusing angles to find because it’s on the ‘outside’ of the angle.

Question 3.
The answer is 90 full turns. What might the question be?
What fraction of a circle is the measurement of 90 in a circle.
Explanation:
Complete circle is divided into 4 equal parts.
Angle of circle is 360 degrees.
Fraction of circle is

Question 4.
Why is a straight angle different from a reflex angle?
A Straight angle or a half turn is always 180 °.
Reflex is the next largest measuring more than 180° but less than 360°.
Explanation:
Full rotation or full circle at 360°.
A reflex angle is always greater than a straight angle which is a half-circle (180°),
and less than a complete angle which is a full circle (360°).
Question 5.
What is the measure of the shaded angle? Through what fraction of the circle does the angle turn?

Estimate.
$$\frac{1}{3}$$ = $$\frac{}{360}$$ 3 × 12 = 36, so 3 × ________ = 360.
The measure of the angle is _________.
$$\frac{1}{3}$$ = $$\frac{}{360}$$ 3 × 12 = 36, so 3 × 120 = 360.
The measure of the angle is 120°
yes, its reasonable.
Explanation:
Circle is divided into 3 equal parts.
Each fraction of a circle is $$\frac{1}{3}$$
Angle of a fraction is $$\frac{360}{3}$$ = 120 degrees.

Tell the measure of the angle in degrees.

Question 6.

The measure of the angle is 120°
Explanation:
whole is divided into 9 equal parts.
fraction of a whole is $$\frac{1}{9}$$ x 360°
= $$\frac{360}{9}$$ = 40°,
So, 9 × 40 =360.
The measure of the angle is 120°

Question 7.

The measure of the angle is 90°
Explanation:
Whole is divided into 4 equal parts,
fraction of a whole is $$\frac{1}{4}$$ x 360°
= $$\frac{360}{4}$$ = 90°,
so 4 × 90 =360.
The measure of the angle is 90°

Question 8.
Why is this angle not a straight angle?

The given angle is more than straight angle and less than a circle.
Explanation:
A reflex angle is any angle that is more than 180 degrees (half circle) and less than 360 degrees (full circle).
A reflex angle will always have either an obtuse or an acute angle on the other side of it.
It can be one of the more confusing angles to find because it’s on the ‘outside’ of the angle.

Classify the angle. Write acute, right, obtuse, straight, or reflex.

Question 9.

Acute angle
Explanation:
An acute angle is defined as an angle that measures less than 90 degrees.

Question 10.

Reflex angle
Explanation:
A reflex angle is any angle that is more than 180 degrees (half circle) and less than 360 degrees (full circle).

Question 11.

Obtuse angle
Explanation:
Obtuse angle is an angle greater than 90°

Question 12.

Straight angle
Explanation:
A straight angle measures 180° it is an angle whose arms lie in opposite directions from the vertex and join together to form 180°.

Question 13.
History The Chicago Cultural Center was completed in 1897 and contains two detailed stained-glass ceiling domes. The Healy and Millet dome is on the north side of the building in the Grand Army of the Republic Hall and was designed and made by two talented artists.

How many parts fill the circle? ________
8 parts

What is the measure of the angle of one fractional part of the stained-glass ceiling? _________
45°
Explanation:
Whole is divided into 8 equal parts,
fraction of a whole is $$\frac{1}{8}$$ x 360°
= $$\frac{360}{8}$$ = 45°,
So, 8 × 45° =360

Question 14.
Open-Ended On a separate piece of paper, design a stained-glass window within a circle that has a center point. Include acute, right, obtuse, straight, and reflex angles.

Explanation:
∠A’OD is an Acute angle — an angle that measures greater than 0 and less than 90°
∠AOB is a Right angle — an angle that measures 90°
∠DOC is a Obtuse angle — an angle that measures greater than 90° and less than 180°
∠A’OA is a Straight angle — an angle that measures 180°
∠AOB’ is a Reflex angle — an angle that measures greater than 180°

I’m in a Learning Mindset!

Are my efforts effective when I write an equation for the degree measure of an angle?


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Rendering an Image of a 3D Scene: an Overview

In the previous chapter, we mentioned that the rendering process could be looked at a two steps process:

• projecting 3D shapes on the surface of a canvas and determining which part of these surfaces are visible from a given point of view,
• simulating the way light propagates through space, which combined with a description of the way light interacts with the materials objects are made of, will give these objects their final appearance (their color, their brightness, their texture, etc.).

In this chapter we will only review the first step in more detail, and more precisely explain how each one of these problems (projecting the objects' shape on the surface of the canvais and the visibility problem) are typically solved. While many solutions may be used, we will only look at the most common ones. This is just an overall presentation. Each method will be studied in a separate lesson and an implementation of these algorithms provided (in a self contained C++ program).

## Going from 3D to 2D: the Projection Matrix

Figure 1: to create an image of a cube, we just need to extend lines from the objects corners towards the eye and find the intersection of these lines with a flat surface (the canvas) perpendicular to the line of sight.

An image is just a representation of a 3D scene on a flat surface: the surface of a canvas or the screen. As explained in the previous chapter, to create an image that looks like reality to our brain, we need to simulate the way an image of the world is formed in our eyes. The principle is quite simple. We just need to extend lines from the objects corners towards the eye and find the intersection of these lines with a flat surface perpendicular to the line of sight. By connecting these points to each other to reform the objects edges, we get a wireframe representation of the scene.

It is important to note, that this sort of construction is in away a completely arbitrary way of flattening a three dimensional world onto a two-dimensional surface. The technique we just described gives us what is called in drawing, a one point perspective projection, and this is generally how we do things in CG because this how the eyes and also cameras work (cameras were obviously designed to produce images similar to the sort of images our eyes create). But in the art world, nothing stops you from coming up with totally different rules. You can in particular get images with several (two, three, four) points perspective.

One of the main important visual properties of this sort of projection is that an object gets smaller as it moves further away from the eye (the rear edges of a box are smaller than the front edges). This effect is called foreshortening.

Figure 2: the line of sight passes through the centre of the canvas.

Figure 3: the size of the canvas can be changed. Making it smaller reduces the field of view.

There's two two important things to note about this type of projection. First the eye is in the centre of the canvas. In other words, the line of sight always passes through the middle of the image (figure 2). Note also that the size of the canvas itself is something we can change. We can more easily understand what the impact of changing the size of the canvas has if we draw the viewing frustum (figure 3). The frustum is the pyramid defined by tracing lines from each corner of the canvas towards the eye, and extending these lines further down into the scene (as a far as the eye can see). It is also referred to as the viewing frustum or viewing volume. You can can easily see that the only objects visible to the camera are those which are contained within the volume of that pyramid. By changing the size of the canvas we can either extend that volume or make it smaller. The larger the volume the more of the scene we see. If you are familiar with the concept of focal length in photography, then you will have recognised that this has the same effect as changing the focal length of photographic lenses. An other way of saying this, is that by changing the size of the canvas, we change the field of view.

Figure 4: when the canvas becomes infinitesimally small, the lines of the frustum become orthogonal to the canvas. When then get what we call an orthographic projection. The game SimCity uses a form of orthographic view which gives it a unique look.

Something interesting happens when when the canvas becomes infinitesimally small: the lines forming the frustum, end up parallel to each other (they are orthogonal to the canvas). This is of course impossible in reality, but not impossible in the virtual world of a computer. In this particular case, you get what we call an orthographic projection. It's important to note that orthographic projection is a form of perspective projection, only one in which the size of the canvas is virtually zero. This has for effect to cancel out the foreshortening effect: the size of the edges of objects are preserved when projected to the screen.

Figure 5: P' is the projection of P on the canvas. The coordinates of P' can easily be computed using the property of similar triangles.

Geometrically, computing the intersection point of these lines with the screen is incredibly simple. If you look at the adjacent figure (where P is the point projected onto the canvas, and P' this project point), you can see that the angle $\angle ABC$ and $\angle AB'C'$ is the same. A is defined as the eye, AB the distance of the point P along the z-axis (P's z-coordinate), and BC the distance of the point P along the y-axis (P's y coordinate). B'C' is the y coordinate of P', and AB' the z-coordinate of P' (and also the distance of the eye to the canvas). When two triangles have the same angle, we say that they are similar. Similar triangles have an interesting property: the ratio of the lengths of their corresponding sides is constant. Based on this property, we can write that: $${ BC \over AB } = { B'C' \over AB' }$$

If we assume that the canvas is located 1 unit away from the eye (in other words that AB' equals 1 (this is purely a convention to simplify this demonstration), and if we substitute AB, BC, AB' and B'C' with their respective points' coordinates, we get: $${ BC \over AB } = { B'C' \over 1 } \rightarrow P'.y = { P.y \over P.z }.$$

In other words, to find the y coordinate of the projected point, you simply need to divide the point y-coordinate by its z-coordinate. The same principle can be used to compute the x coordinate of P':

$$P'.x = { P.x \over P.z }.$$

This is a very simple and yet this is an extremely important relationship in computer graphics, known as the perspective divide or z-divide (if you were on desert island and needed to remember something about computer graphics, that would probably be this equation).

In computer graphics, we generally perform this operation using what we call a perspective projection matrix. As its name indicates, it's a matrix which when applied to points, project them to the screen. In the next lesson, we will explain step by step how and why this matrix works, learn how to build it and use it.

But wait! The problem is that whether you need the perspective projection totally depends on the technique you use to sort out the visibility problem. Anticipating what we will learn in the second part of this chapter, algorithms for solving the visibility problems comes into two main categories:

• Rasterisation,
• Ray-tracing.

Algorithms of the first category, relies on projecting P onto the screen to compute P'. For these algorithms, the perspective projection matrix is therefore needed. In ray tracing, rather than projecting the geometry onto the screen, we trace a ray passing through P' and look for P. Obviously we don't need to project P anymore with this approach since we already know P', which means that in ray tracing, the perspective projection is actually technically not needed (and therefore never used).

We will study the two algorithms in detail in the next chapters and the next lessons. However it is important to understand the difference between the two and how they work at this point. As explained before, the geometry needs to be projected onto the surface of the canvas. To do so, P is projected along an "implicit" line (implicit because we actually never really need to build this line as we need to with ray tracing) connecting P to the eye. You can see the process as if you were moving a point along that line from P to the eye, until it lies onto the canvas. That point would be P'. In this approach you know P, but you don't know P'. You compute it using the projection approach. But you can also look at the problem the other way around. You can wonder whether for any point on the canvas (say P' - which by default we will assume is in the centre of the pixel), there is a point P on the surface of the geometry that projects onto P'. The solution to this problem, is to actually explicitly this time create a ray from the eye to P', extend or project this ray down into the scene, and find out if this ray intersects any 3D geometry. If if does, then the intersection point is P. Hopefully, you can now see more distinctively the difference between rasterisation (we know P, we compute P') and ray tracing (we know P', we look for P).

The advantage of the rasterisation approach over ray tracing is mainly speed. Computing the intersection of rays with geometry is a computationally expensive operation. This intersection time also grows linearly with the amount of geometry contained in the scene, as we will see in one of the next lessons. On the other hand, the projection process is incredibly simple, relies on basic math operations (multiplications, divisions, etc.), and can be aggressively optimised (especially if special hardware is designed for this purpose which is the case with GPUs). Graphics cards are almost all using an algorithm based on the rasterisation approach (which is one of the reasons they can render 3D scene so quickly, at interactive frame rates). When realtime rendering APIs such as OpenGL or DirectX are used, the projection matrix needs to be dealt with. Even if you are only interested in ray tracing, you should know about it for at least historical reason: it is one of the most important techniques in rendering and the most commonly used technique for producing real-time 3D computer graphics. Plus, it is likely at some point that you will have to deal with the GPU anyway, and real time rendering APIs do not compute this matrix for you. You will have to do it yourself.

The concept of rasterisation is really important in rendering. As we learned in this chapter, the projection of P onto the screen, can be computed by dividing the point's coordinates x and y by the point's z-coordinate. As you may guess, all initial coordinates are real numbers - floats for instance - thus P' coordinates are also real numbers. However pixel coordinates need to be integers, thereby, to store the color of P's in the image, we will need to convert its coordinates to pixel coordinates - in other words from floats to integers. We say that the points coordinates are converted from screen space to raster space. More information can be found on this process in the lesson on rays and cameras.

The next three lessons are devoted to studying the construction of the orthographic and perspective matrix, and how to use them in OpenGL to display images and 3D geometry.


text:
# discretize.TreeMesh.get_face_inner_product¶

TreeMesh.get_face_inner_product(model=None, invert_model=False, invert_matrix=False, do_fast=True, **kwargs)[source]

Generate the face inner product matrix or its inverse.

This method generates the inner product matrix (or its inverse) when discrete variables are defined on mesh faces. It is also capable of constructing the inner product matrix when physical properties are defined in the form of constitutive relations. For a comprehensive description of the inner product matrices that can be constructed with get_face_inner_product, see Notes.

Parameters
modelNone or numpy.ndarray, optional

Parameters defining the material properties for every cell in the mesh. Inner product matrices can be constructed for the following cases:

• None : returns the basic inner product matrix

• (n_cells) numpy.ndarray : returns inner product matrix for an isotropic model. The array contains a scalar physical property value for each cell.

• (n_cells, dim) numpy.ndarray : returns inner product matrix for diagonal anisotropic case. Columns are ordered np.c_[σ_xx, σ_yy, σ_zz]. This can also a be a 1D array with the same number of total elements in column major order.

• (n_cells, 3) numpy.ndarray (dim is 2) or (n_cells, 6) numpy.ndarray (dim is 3) : returns inner product matrix for full tensor properties case. Columns are ordered np.c_[σ_xx, σ_yy, σ_zz, σ_xy, σ_xz, σ_yz] This can also be a 1D array with the same number of total elements in column major order.

invert_modelbool, optional

The inverse of model is used as the physical property.

invert_matrixbool, optional

Returns the inverse of the inner product matrix. The inverse not implemented for full tensor properties.

do_fastbool, optional

Do a faster implementation (if available).

Returns
(n_faces, n_faces) scipy.sparse.csr_matrix

inner product matrix

Notes

For continuous vector quantities $$\vec{u}$$ and $$\vec{w}$$ whose discrete representations $$\mathbf{u}$$ and $$\mathbf{w}$$ live on the faces, get_face_inner_product constructs the inner product matrix $$\mathbf{M_\ast}$$ (or its inverse $$\mathbf{M_\ast^{-1}}$$) for the following cases:

Basic Inner Product: the inner product between $$\vec{u}$$ and $$\vec{w}$$

$\langle \vec{u}, \vec{w} \rangle = \mathbf{u^T \, M \, w}$

Isotropic Case: the inner product between $$\vec{u}$$ and $$\sigma \vec{w}$$ where $$\sigma$$ is a scalar function.

$\langle \vec{u}, \sigma \vec{w} \rangle = \mathbf{u^T \, M_\sigma \, w}$

Tensor Case: the inner product between $$\vec{u}$$ and $$\Sigma \vec{w}$$ where $$\Sigma$$ is tensor function; $$\sigma_{xy} = \sigma_{xz} = \sigma_{yz} = 0$$ for diagonal anisotropy.

$\begin{split}\langle \vec{u}, \Sigma \vec{w} \rangle = \mathbf{u^T \, M_\Sigma \, w} \;\;\; \textrm{where} \;\;\; \Sigma = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{bmatrix}\end{split}$

Examples

Here we provide some examples of face inner product matrices. For simplicity, we will work on a 2 x 2 x 2 tensor mesh. As seen below, we begin by constructing and imaging the basic face inner product matrix.

>>> from discretize import TensorMesh
>>> import matplotlib.pyplot as plt
>>> import numpy as np
>>> import matplotlib as mpl

>>> h = np.ones(2)
>>> mesh = TensorMesh([h, h, h])
>>> Mf = mesh.get_face_inner_product()

>>> fig = plt.figure(figsize=(6, 6))
>>> ax = fig.add_subplot(111)
>>> ax.imshow(Mf.todense())
>>> ax.set_title('Basic Face Inner Product Matrix', fontsize=18)
>>> plt.show()

Next, we consider the case where the physical properties of the cells are defined by consistutive relations. For the isotropic, diagonal anisotropic and full tensor cases, we show the physical property tensor for a single cell.

Define 4 constitutive parameters and define the tensor for each cell for isotropic, diagonal and tensor cases.

>>> sig1, sig2, sig3, sig4, sig5, sig6 = 6, 5, 4, 3, 2, 1
>>> sig_iso_tensor = sig1 * np.eye(3)
>>> sig_diag_tensor = np.diag(np.array([sig1, sig2, sig3]))
>>> sig_full_tensor = np.array([
...     [sig1, sig4, sig5],
...     [sig4, sig2, sig6],
...     [sig5, sig6, sig3]
... ])

Then plot matrix entries,

Expand to show scripting for plot
>>> fig = plt.figure(figsize=(15, 5))
>>> ax1 = fig.add_subplot(131)
>>> ax1.imshow(sig_iso_tensor)
>>> ax1.axis('off')
>>> ax1.set_title("Tensor (isotropic)", fontsize=16)
>>> ax2 = fig.add_subplot(132)
>>> ax2.imshow(sig_diag_tensor)
>>> ax2.axis('off')
>>> ax2.set_title("Tensor (diagonal anisotropic)", fontsize=16)
>>> ax3 = fig.add_subplot(133)
>>> ax3.imshow(sig_full_tensor)
>>> ax3.axis('off')
>>> ax3.set_title("Tensor (full anisotropic)", fontsize=16)
>>> plt.show()

(png, pdf)

Here, construct and image the face inner product matrices for the isotropic, diagonal anisotropic and full tensor cases. Spy plots are used to demonstrate the sparsity of the inner product matrices.

Isotropic case:

>>> v = np.ones(mesh.nC)
>>> sig = sig1 * v
>>> M1 = mesh.get_face_inner_product(sig)

Diagonal anisotropic case:

>>> sig = np.c_[sig1*v, sig2*v, sig3*v]
>>> M2 = mesh.get_face_inner_product(sig)

Full anisotropic case:

>>> sig = np.tile(np.c_[sig1, sig2, sig3, sig4, sig5, sig6], (mesh.nC, 1))
>>> M3 = mesh.get_face_inner_product(sig)

And then we can plot the sparse representation,

Expand to show scripting for plot
>>> fig = plt.figure(figsize=(12, 4))
>>> ax1 = fig.add_subplot(131)
>>> ax1.spy(M1, ms=5)
>>> ax1.set_title("M (isotropic)", fontsize=16)
>>> ax2 = fig.add_subplot(132)
>>> ax2.spy(M2, ms=5)
>>> ax2.set_title("M (diagonal anisotropic)", fontsize=16)
>>> ax3 = fig.add_subplot(133)
>>> ax3.spy(M3, ms=5)
>>> ax3.set_title("M (full anisotropic)", fontsize=16)
>>> plt.show()

(png, pdf)


text:
# Is Gaussian Process Regression a linear model?

I had a discussion today with someone saying that Gaussian Processes are linear models. I don't see in which sense this may be correct. To be clear, here the definition of a linear model is the usual one, i.e., a model which is linear in the parameters. Thus,

$y=\beta_0+\beta_1x+\beta_2\sin(x)+\epsilon$

and

$y=\beta_0+\boldsymbol{\beta}^T\cdot\mathbf{x}+\epsilon$

are linear, and

$y=\beta_0+\beta_1x+\beta_2\exp({\beta_3x})+\epsilon$

is not.

For simplicity, let's consider a Squared Exponential covariance function, and assume that the correlation length, the signal variance and the noise variance are known. Given a design matrix $X$ and corresponding response vector $\mathbf{y}$, the GP prediction at a new prediction point $\mathbf{x}^*$ is

$$\hat{y}(\mathbf{x}^*)=\mathbf{k}_*^T(K+\sigma I)^{-1}\mathbf{y}$$

Now, this estimator is clearly a nonlinear function of $X$ and a linear function of $\mathbf{y}$. The other person insisted that $\mathbf{y}$ is the parameter vector of this model, and thus the model is linear. I don't think this makes any sense: it would mean that the number of parameters of the model depends on the sample size. I think we can at most say that the estimator is a linear function of $\mathbf{y}$, but surely not the statistical model underlying Gaussian Process Regression. Do you agree?

• Have you noticed that the OLS estimator $(X^\prime X)^{-}X^\prime y$ is highly nonlinear in $X$, too? – whuber Oct 31 '17 at 19:43
• @whuber sure, and it's also linear in $\mathbf{y}$, like the GPR estimator. However, I was talking about linearity of the model, not the estimator. I guess the other person confounded the two things. – DeltaIV Oct 31 '17 at 21:58
• Interestingly, a GP regression model is equivalent to (linear) ridge regression if you use a linear kernel. – Kevin Yang Nov 1 '17 at 1:13
• @KevinYang true, but a linear kernel is a degenerate kernel, in that its eigenexpansion is finite (and thus the covariance function is only semidefinite positive, instead than definite positive). Let's consider just strictly positive definite covariance functions: do you think GP regression is a linear model or not? If it is, and you can show me which are the model parameters I would accept it as an answer. However, if the parameters depend on the data set, I don't think we can talk of a linear model...ideally the model is defined independently of the data set. – DeltaIV Nov 1 '17 at 11:25

I think the technically correct term to use here is that GP regression is a linear smoother, i.e. its predictions are a linearly weighted combination of past observed outputs. This does not make the model as such linear. For that to be true, the predictions must be a linear function of the inputs. This is only the case with GPs if you use a linear covariance function.

The statistical definition of a model being linear is that the model must be linear in its parameters. Gaussian Process Regression can be defined by using either the function-space view or the weight-space view to reach the formula for the posterior mean and posterior variance.

If we see the weight-space view, we can clearly see that Gaussian Process Regression is indeed a linear model with non linear functions of the inputs. The model is defined as a bayesian linear regression model:

$$f(\mathbf{x}) = \phi(\mathbf{x})^T \mathbf{w}$$ $$and$$ $$y = f(\mathbf{x}) + \epsilon$$ where:

• $$\mathbf{x}$$ denotes the input vector for an input

• $$\phi(X)$$ denotes some basis function applied on the input space

• $$\mathbf{w}$$ is the weight vector

• (other symbols defined as usual).

However non-linearity of inputs doesn't affect the linearity of the model itself. Kindly refer to the derivation in the weight-space view section of http://www.gaussianprocess.org/gpml/chapters/RW2.pdf for more information.


text:
# ParametricPlot on Piecewise function not working with consecutive t values [closed]

I am trying to graph a piecewise function composed of a series of parametric cubic functions. I initially created a ParametricPlot for the first four pieces of the equation and it was working fine, as shown below

ParametricPlot[
Piecewise[
{{{1.27 t^3 - 1.57 t + 0.62, -3.5 t^3 - 0.54 t + 0.76},
0 < t <= 0.06},
{{-0.62t^3 + 0.35t^2 - 1.59t + 0.62, 2.04t^3 - 1.03t^2 - 0.47t + 0.76},
0.06 < t <= 0.15},
{{1.49t^3 - 0.57t^2 - 1.45t + 0.61 ,-3.47t^3 + 1.39t^2 - 0.83t + 0.78  },
0.15 < t <= 0.31},
{{14.35t^3 - 12.51t^2 + 2.24t + 0.23 , -3.89t^3 + 1.77t^2 - 0.95t + 0.79},
0.31 < t <= .41}}],
{t, 0, .41}]

outputting this graph

However, once I added the next term and adjusted the ParametricPlot accordingly, it no longer worked.

The code:

ParametricPlot[
Piecewise[
{{{1.27 t^3 - 1.57 t + 0.62, -3.5 t^3 - 0.54 t + 0.76},
0 < t <= 0.06},
{{-0.62t^3 + 0.35t^2 - 1.59t + 0.62, 2.04t^3 - 1.03t^2 - 0.47t + 0.76},
0.06 < t <= 0.15},
{{1.49t^3 - 0.57t^2 - 1.45t + 0.61 ,-3.47t^3 + 1.39t^2 - 0.83t + 0.78  },
0.15 < t <= 0.31},
{{14.35t^3 - 12.51t^2 + 2.24t + 0.23 , -3.89t^3 + 1.77t^2 - 0.95t + 0.79},
0.31 < t <= .41},
{{-5.93t^3 + 12.45t^2 - 8t + 1.63 ,13.95t^3 - 20.19t^2 + 8.07t - 0.44}
0.41 < t <= 0.49},
{{6.62t^3 - 5.91t^2 + 0.94t + 0.18 , 2.96t^3 - 4.12t^2 + 0.23t + 0.83}
0.49 < t <= 0.53}}}],
{t,0,.53}]

with the following error message. The first one is repeated twice.

General: Further output of Piecewise::pairs will be suppressed during this calculation.

While it seems that the issue is that the data is not in pairs, I do not understand why it worked for the first part and how to get it to continue working as I add more data. Did I just screw up putting it in?

• You have made a simple typing mistake: you are omitting two commas from the last two segments that you added in 2nd example. You need commas before 0.41 < t <= 0.49 and 0.49 < t <= 0.53. – m_goldberg Jan 11 at 18:13

This is just to fill out what I already said in a comment to the question.

You have made a simple typing mistake. You are omitting two commas from the last two segments that you added in 2nd example. You need commas before 0.41 < t <= 0.49 and 0.49 < t <= 0.53 Here is how plot looks when those commas are inserted:

ParametricPlot[
Piecewise[
{{{1.27 t^3 - 1.57 t + 0.62, -3.5 t^3 - 0.54 t + 0.76},
0 < t <= 0.06},
{{-0.62 t^3 + 0.35 t^2 - 1.59 t + 0.62, 2.04 t^3 - 1.03 t^2 - 0.47 t + 0.76},
0.06 < t <= 0.15},
{{1.49 t^3 - 0.57 t^2 - 1.45 t + 0.61, -3.47 t^3 + 1.39 t^2 - 0.83 t + 0.78},
0.15 < t <= 0.31},
{{14.35 t^3 - 12.51 t^2 + 2.24 t + 0.23, -3.89 t^3 + 1.77 t^2 - 0.95 t + 0.79},
0.31 < t <= .41},
{{-5.93 t^3 + 12.45 t^2 - 8 t + 1.63, 13.95 t^3 - 20.19 t^2 + 8.07 t - 0.44},
0.41 < t <= 0.49},
{{6.62 t^3 - 5.91 t^2 + 0.94 t + 0.18, 2.96 t^3 - 4.12 t^2 + 0.23 t + 0.83},
0.49 < t <= 0.53}}],
{t, 0, .53}]