Section2.6Trigonometric Functions

Subsection2.6.1Transformation of Trigonometric Functions

Recall the graphs of the six basic trigonometric functions reviewed in Section 1.3.4:

We will now look at a specific mathematical model of a phenomenon exhibiting cyclical behaviour — the so-called predator-prey population model.

Example2.25. The Predator-Prey Population Model.

The population of owls (predators) in a certain region over a $2$-year period is estimated to be

\begin{equation*} P(t)=1000+100\sin\left(\frac{\pi t}{12}\right) \end{equation*}

in month $t\text{,}$ and the population of mice (prey) in the same area at time $t$ is given by

\begin{equation*} p(t)=20,000 + 4000\cos\left(\frac{\pi t}{12}\right)\text{.} \end{equation*}

Sketch the graphs of these two functions and explain the relationship between the sizes of the two populations.

Solution

We first observe that both of the given functions are periodic with period $24$ months. To see this, recall that both the sine and cosine functions are periodic with period $2\pi\text{.}$ The smallest value of $t > 0$ such that $\sin\left(\pi t/12\right)=0$ is then obtained by solving the equation

\begin{equation*} t = \frac{2\pi}{\pi/12}\text{,} \end{equation*}

giving $t=24$ as the period of $\sin\left(\pi t/12\right)\text{.}$ Since $P(t+24)=P(t)\text{,}$ we see that the function $P$ is periodic with period $24\text{.}$ Similarly, one verifies that the function $p$ is also periodic with period $24\text{.}$ Next, recall that both the sine and cosine functions oscillate between $-1$ and $+1$ so that $P(t)$ is seen to oscillate between $\left[1000+100(-1)\right]\text{,}$ or $900\text{,}$ and $\left[1000+100(1)\right]\text{,}$ or $1100\text{,}$ while $p(t)$ oscillates between $\left[20,000+4000(-1)\right]\text{,}$ or $16,000\text{,}$ and $\left[20,000+4000(1)\right]\text{,}$ or $24,000\text{.}$ Finally, plotting a few points on each graph for –say, $t=0,2,3\text{,}$ and so on — we obtain the graphs of the functions $P$ and $p$ as shown below.

From the graphs, we see that at time $t=0\text{,}$ the predator population stands at $1000$ owls. As the predator population increases, the prey population decreases from $24,000$ at that instant. Eventually, this decrease in the food supply causes the predator population to decrease, which in turn allows for an increase in the prey population. But as the prey population increases, resulting in an increase in food supply, the predator population once again increases. The cycle is complete and starts all over again.

For the solution of Example 2.25, we described how to plot sine and cosine functions when they have been transformed by some translations,reflections, and stretchings in the horizontal and vertical directions. Recall from Section 2.2 that this has already been described for functions in general. However, since trigonometric functions are special in that they have periodic behaviour, we will summarize transformations of sine and cosine functions. This information will help when graphing these functions in general, if we know the graphs of the basic functions $y=\sin x$ and $y= \cos x\text{.}$ The tangent function can be dealt with in a similar manner; just remember that it has period $\pi\text{.}$

Transformation of the Sine and Cosine Functions.

Given the function

\begin{equation*} F(x)=Af(B(x-C))+D \end{equation*}

with real constants $A\neq 0, B \neq 0, C$ and $D\text{,}$ and $f(x) = \sin x$ or $f(x) = \cos x\text{,}$ then the graph of $F$ is transformed as follows from $f\text{:}$

• The constant $\lvert A \rvert$ is called the amplitude and stretches $f$ vertically. Furthermore, if $A \lt 0$ then $f$ reflects in the $x$-axis.

• The constant $\frac{2\pi}{\lvert B \rvert}$ is called the period and stretches $f$ horizontally. Furthermore, if $B \lt 0$ then $f$ reflects in the $y$-axis.

• If $C > 0$ then $f$ shifts to the right, otherwise to the left.

• If $D > 0$ then $f$ shifts up, otherwise down.

Interactive Demonstration. Investigate the transformations $y = Af(B(x-C))+D$ with the graph of $f(x) = \sin(x)$ below:

Transformation of the Tangent Function.

Given the function

\begin{equation*} F(x)=Af(B(x-C))+D \end{equation*}

with real constants $A\neq 0, B \neq 0, C$ and $D\text{,}$ and $f(x) = \tan x\text{,}$ then the graph of $F$ is transformed as follows from $f\text{:}$

• The constant $\lvert A \rvert$ stretched $f$ vertically and in this case is not an amplitude. Furthermore, if $A \lt 0$ then $f$ reflects in the $x$-axis.

• The constant $\frac{\pi}{\lvert B \rvert}$ is called the period and stretches $f$ horizontally. Furthermore, if $B \lt 0$ then $f$ reflects in the $y$-axis.

• If $C > 0$ then $f$ shifts to the right, otherwise to the left.

• If $D > 0$ then $f$ shifts up, otherwise down.

Note: Remember the order in which to perform the translations to produce the correct graph: first reflections and stretchings, then translations.
Example2.26. Transformation of Cosine.

Graph the function defined by $f(x)=3\cos(2x)+4\text{.}$

Solution

The graph of the basic function $y=\cos(x)$ is translated vertically up by $4$ units:

The amplitude of this cosine function is $3\text{,}$ which means we now stretch the graph of the cosine function vertically by $3$ units:

Lastly, the period is $\frac{2\pi}{2} = \pi\text{,}$ which means the graph of the cosine function is compressed by a factor of $\frac{1}{2}\text{:}$

This results in the graph of $f$ as desired (in blue). There are no horizontal or vertical reflections, since both $3>0$ and $2>0\text{.}$

Subsection2.6.2Inverse Trigonometric Functions

The trigonometric functions frequently arise in problems, and often it is necessary to invert the functions, for example, to find an angle with a specified sine. Of course, there are many angles with the same sine, so the sine function doesn't actually have an inverse that reliably “undoes” the sine function. If you know that $\sin x=0.5\text{,}$ you can't reverse this to discover $x\text{,}$ that is, you can't solve for $x\text{,}$ as there are infinitely many angles with sine $0.5\text{.}$ Nevertheless, it is useful to have something like an inverse to the sine, however imperfect. The usual approach is to pick out some collection of angles that produce all possible values of the sine exactly once. If we “discard” all other angles, the resulting function does have a proper inverse.

The sine takes on all values between $-1$ and $1$ exactly once on the interval $[-\pi/2,\pi/2]\text{.}$

If we truncate the sine, keeping only the interval $[-\pi/2,\pi/2]\text{,}$ then this truncated sine has an inverse function. We call this the inverse sine or the arcsine, and write it in one of two common notation: $y=\arcsin(x)\text{,}$ or $y=\sin^{-1}(x)\text{.}$

Definition2.27. Inverse Sine Function.

Given the restricted sine funtion $y=\sin(x)$ with domain $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{,}$ the inverse sine function (or arcsine function) is denoted by

\begin{equation*} y=\sin^{-1}(x) \qquad (\text{ or, } y=\arcsin(x))\text{,} \end{equation*}

with domain $\left[-1,1\right]$ and range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{.}$

Recall that a function and its inverse undo each other in either order, for example, $\ds (\root3\of x)^3=x$ and $\ds \root3\of{x^3}=x\text{.}$ This does not work with the sine and the “inverse sine” because the inverse sine is the inverse of the truncated sine function, not the real sine function. It is true that $\sin(\arcsin(x))=x\text{,}$ that is, the sine undoes the arcsine. It is not true that the arcsine undoes the sine, for example, $\sin(5\pi/6)=1/2$ and $\arcsin(1/2)=\pi/6\text{,}$ so doing first the sine then the arcsine does not get us back where we started. This is because $5\pi/6$ is not in the domain of the truncated sine. If we start with an angle between $-\pi/2$ and $\pi/2$ then the arcsine does reverse the sine: $\sin(\pi/6)=1/2$ and $\arcsin(1/2)=\pi/6\text{.}$

Example2.28. Arcsine of Common Values.

Compute $\sin^{-1}(0)\text{,}$ $\sin^{-1}(1)$ and $\sin^{-1}(-1)\text{.}$

Solution

These come directly from the graph of $y=\arcsin x\text{:}$

We can do something similar for the cosine function. As with the sine, we must first truncate the cosine so that it can be inverted, in particular, we use the interval $[0,\pi]\text{.}$

Note that the truncated cosine uses a different interval than the truncated sine, so that if $y=\arccos(x)$ we know that $0\le y\le \pi\text{.}$

Example2.29. Arccosine of Common Values.

Compute $\cos^{-1}(0)\text{,}$ $\cos^{-1}(1)$ and $\cos^{-1}(-1)\text{.}$

Solution

These come directly from the graph of $y=\arccos x\text{:}$

The truncated tangent uses an interval of $(-\pi/2,\pi/2)\text{.}$

Reflecting the truncated tangent in the line $y=x$ gives the arctangent function.

Example2.30. Arctangent of Common Values.

Compute $\tan^{-1}(0)\text{.}$ What value does $\tan^{-1}x$ approach as $x$ gets larger and larger? What value does $\tan^{-1}x$ approach as $x$ gets large (and negative)?

Solution

These come directly from the graph of $y=\arctan x\text{.}$ In particular, $\tan^{-1}(0)=0\text{.}$ As $x$ gets larger and larger, $\tan^{-1}x$ approaches a value of $\frac{\pi}{2}\text{,}$ whereas, as $x$ gets large but negative, $\tan^{-1}x$ approaches a value of $-\frac{\pi}{2}\text{.}$

The following definition summarizes the inverse trigonometric functions and their respective domains and restricted ranges.

Definition2.31. Inverses of the Primary Trigonometric Functions.
\begin{equation*} \begin{array}{llll} \text{ notation: } \amp y=\sin^{-1}(x) \amp y=\cos^{-1}(x) \amp y = \tan^{-1}(x) \\ \text{ domain: } \amp \left[-1,1\right] \amp \left[-1,1\right] \amp \left(-\infty,\infty\right) \\ \text{ range: } \amp \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \amp \left[0,\pi\right] \amp \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \end{array} \end{equation*}

While the inverses for the sine, cosine and tangent are enough for most purposes, we state the remaining trigonometric inverses for completeness.

Definition2.32. Inverses of the Secondary Trigonometric Functions.
\begin{equation*} \begin{array}{llll} \text{ notation: } \amp y=\csc^{-1}(x) \amp y=\sec^{-1}(x) \amp y = \cot^{-1}(x) \\[1ex] \text{ domain: } \amp \left(-\infty,-1\right]\cup\left[-1,\infty\right) \amp \left(-\infty,-1\right]\cup\left[1,\infty\right) \amp \left(-\infty,\infty\right) \\[1ex] \text{ range: } \amp \left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right] \amp \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right] \amp \left(0,\pi\right)\\ ~ \amp \end{array} \end{equation*}

Note:

1. We use both $\sin^{-1}(x)$ and $\arcsin(x)$ to represent inverse sine.

2. $\sin^{-1}(x) \neq (\sin x)^{-1} \neq \sin(x^{-1})$ since $\sin^{-1}(x)$ denotes the inverse sine function, $(\sin x)^{-1} = \frac{1}{\sin x}$ denotes the reciprocal of sine, and $\sin x^{-1} = \sin \frac{1}{x}$ denotes the sine of the reciprocal of $x\text{.}$

The Cancellation Rules are tricky since we restricted the domains of the trigonometric functions in order to obtain inverse trig functions:

Cancellation Rules.
\begin{equation*} \sin ( \sin^{-1} x ) = x, \ \ \ x \in [-1, 1] \qquad\qquad \sin^{-1} ( \sin x) = x, \ \ \ x \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right] \end{equation*}
\begin{equation*} \cos ( \cos^{-1} x ) = x, \ \ \ x \in [-1, 1] \qquad\qquad \ \ \ \cos^{-1} ( \cos x) = x, \ \ \ x \in \left[0, \pi \right]\ \ \ \end{equation*}
\begin{equation*} \tan ( \tan^{-1} x ) = x, \ \ \ x \in (-\infty, \infty) \qquad\qquad \tan^{-1} ( \tan x) = x, \ \ \ x \in \left(-\frac{\pi}{2}, \frac{\pi}{2} \right) \end{equation*}
Example2.33. Arcsine.

Find $\sin ^{-1}\left( 1/2\right)\text{.}$

Solution

Since $\sin^{-1}(x)$ outputs values in $[-\pi/2,~\pi/2]\text{,}$ the answer must be in this interval. Let $\theta=\sin^{-1}(1/2)\text{.}$ We need to compute $\theta\text{.}$ Take the sine of both sides to get $\sin \theta = \sin(\sin^{-1}(1/2))=1/2$ by the Cancellation Rule. There are many angles $\theta$ that work, but we want the one in the interval $[-\pi/2,~\pi/2]\text{.}$ Thus, $\theta=\pi/6$ and hence, $\ds\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\text{.}$

Example2.34. Arccosine.

Find $\cos^{-1}\left(\frac{\sqrt 3}{2}\right)\text{.}$

Solution

Trick: Let $\theta=\cos^{-1}\left(\frac{\sqrt 3}{2}\right)\text{.}$ Need to compute $\theta\text{.}$

Take $\cos$ of both sides:

\begin{equation*} \cos \theta = \cos\left(\cos^{-1}\left(\frac{\sqrt 3}{2}\right)\right)=\frac{\sqrt{3}}{2} \text{ by the Cancellation Rule. } \end{equation*}

There are many angles $\theta$ that work, but we need the one in the interval $[0,~\pi]\text{.}$

By the special triangle shown and the definition of cosine:

\begin{equation*} \theta = \frac{\pi}{6} \text{ and so } \cos^{-1}\left(\frac{\sqrt 3}{2}\right) =\frac{\pi}{6}\text{.} \end{equation*}
Example2.35. Arctangent.

Find $\tan^{-1}(\sqrt 3)\text{.}$

Solution

Trick: Let $\theta=\tan^{-1}(\sqrt 3)\text{.}$ Need to compute $\theta\text{.}$

Take $\tan$ of both sides:

\begin{equation*} \tan \theta = \tan(\tan^{-1}(\sqrt 3)=\sqrt 3 \text{ by the Cancellation Rule. } \end{equation*}

There are many angles $\theta$ that work, but we need the one in the interval $(-\pi/2,~\pi/2)\text{.}$

Thus, $\theta=\pi/3$ by the special triangle shown and the definition of tangent:

\begin{equation*} \theta = \frac{\pi}{3}\text{ and so } \tan^{-1}(\sqrt 3)=\frac{\pi}{3}\text{.} \end{equation*}
Example2.36. Cancellation Rule.

Find $\cos^{-1}\left( \cos \left(-5\pi /3\right) \right)\text{.}$

Solution

Range: $\cos^{-1}(x)$ outputs values in $[0,~\pi]\text{,}$ thus the answers must be in this interval.

We can not cancel yet. Instead, we add/subtract multiples of $2\pi$ until we get a number in this range.

By the periodicity of $\cos x\text{,}$ we have

\begin{equation*} \cos \left( -\dfrac{5\pi }{3}\right) =\cos \left( 2\pi -\dfrac{5\pi }{3}\right) =\cos \left( \dfrac{\pi }{3}\right)\text{.} \end{equation*}

By the Cancellation Rule:

\begin{equation*} \cos ^{-1}\left( \cos \left( -\dfrac{5\pi }{3}\right) \right) =\cos^{-1}\left( \cos \left( \frac{\pi }{3}\right) \right) =\frac{\pi }{3}\text{.} \end{equation*}
Example2.37. Arccosine and the Cancellation Rule.

Compute $\cos ^{-1}(\cos(0))\text{,}$ $\cos ^{-1}(\cos(\pi))\text{,}$ $\cos^{-1}(\cos(2\pi))\text{,}$ $\cos^{-1}(\cos(3\pi))\text{.}$

Solution

Since $\cos^{-1}(x)$ outputs values in $[0,~\pi]\text{,}$ the answers must be in this interval. The first two we can cancel using the Cancellation Rules:

The third one we cannot cancel since $2\pi\notin[0,\pi]\text{:}$

\begin{equation*} \cos ^{-1}(\cos(2\pi))\mbox{is NOT equal to } 2\pi\text{.} \end{equation*}

But we know that cosine is a $2\pi$-periodic function, so $\cos(2\pi)=\cos(0)\text{:}$

\begin{equation*} \cos ^{-1}(\cos(2\pi))=\cos^{-1}(\cos(0))=0 \end{equation*}

Similarly with the fourth one, we can \emph{not} cancel yet since $3\pi\notin[0,\pi]\text{.}$ Using $\cos(3\pi)=\cos(3\pi-2\pi)=\cos(\pi)\text{:}$

\begin{equation*} \cos ^{-1}(\cos(3\pi))=\cos^{-1}(\cos(\pi))=\pi\text{.} \end{equation*}
Example2.38. The Triangle Technique I.

Rewrite the expression $\cos(\sin^{-1}x)$ without trig functions. Note that the domain of this function is all $x\in[-1,1]\text{.}$

Solution

Let $\theta=\sin^{-1}x\text{.}$ We need to compute $\cos\theta\text{.}$ Taking the sine of both sides gives $\sin \theta = \sin(\sin^{-1}(x))=x$ by the Cancellation Rule. We then draw a right triangle using $\sin\theta=x/1\text{:}$

If $z$ is the remaining side, then by the Pythagorean Theorem:

and hence $z=+\sqrt{1-x^2}$ since $\theta\in[-\pi/2,~\pi/2]\text{.}$ Thus,

\begin{equation*} \ds\cos\theta=\sqrt{1-x^2}, \end{equation*}

so,

\begin{equation*} \ds\cos(\sin^{-1}x)=\sqrt{1-x^2}\text{.} \end{equation*}
Example2.39. The Triangle Technique II.

For $x\in(0,1)\text{,}$ rewrite the expression $\ds\sin(2\cos^{-1}x)\text{.}$ Compute $\sin(2\cos^{-1}(1/2))\text{.}$

Solution

Let $\theta=\cos^{-1}x$ so that $\cos\theta = x\text{.}$ The question now asks for us to compute $\sin(2\theta)\text{.}$ We then draw a right triangle using $\cos\theta=x/1\text{:}$

To find $\sin(2\theta)$ we use the double angle formula

\begin{equation*} \ds \sin(2\theta)=2\sin\theta\cos\theta\text{.} \end{equation*}

But $\sin\theta=\sqrt{1-x^2}\text{,}$ for $\theta\in[0,\pi]\text{,}$ and $\cos\theta=x\text{.}$ Therefore,

\begin{equation*} \ds\sin(2\cos^{-1}x)=2x\sqrt{1-x^2}\text{.} \end{equation*}

When $x=1/2$ we have

\begin{equation*} \ds\sin(2\cos^{-1}(1/2))=\frac{\sqrt 3}{2}\text{.} \end{equation*}
Exercises for Section 2.6.

Use transformations to sketch the graphs of the following functions.

1. $y=\sin(2x)$ over the interval $\left[0,2\pi\right]$

2. $y=-\sin(x)$ over the interval $\left[0,2\pi\right]$

3. $y=3\sin(4x)-5$ over the interval $\left[0,2\pi\right]$

Solution

We wish to sketch the graph of $y=3\sin(4x)-5$ on the interval $\left[0,2\pi\right]\text{.}$ We first sketch $y=\sin(4x)\text{,}$ which has a period of $2\pi / 4 = \pi/2\text{.}$

Next, we stretch the graph vertically so that the resulting function has an amplitude of 3.
Finally, we shift the graph downwards by 5 units.

4. $y=2\sin\left(x-\frac{\pi}{2}\right)+1$ over the interval $\left[0,2\pi\right]$

5. $y=\tan\left(\frac{x}{3}\right)$ over the interval $\left[-2\pi,2\pi\right]$

6. $y=-\tan(x)+2$ over the interval $\left[-2\pi,2\pi\right]$

7. $y=\tan\left(x-\frac{\pi}{2}\right)$ over the interval $\left[-2\pi,2\pi\right]$

Sales of large kitchen appliances such as ovens and fridges are usually subject to seasonal fluctuations. Everything Kitchen's sales of fridge models from the beginning of $2001$ to the end of $2002$ can be approximated by

\begin{equation*} S(x)=\frac{1}{10}\sin\left(\frac{\pi}{2}(x+1)\right)+\frac{1}{2} \end{equation*}

where $x$ is time in quarters, $x=1$ represents the end of the first quarter of $2001\text{,}$ and $S$ is measured in millions of dollars.

1. What are the maximum and minimum quarterly revenues for fridges?

$$600,000\text{;}$$$400,000\text{.}$
Solution

For any $f(x)\text{,}$ we have that $\sin(f(x))$ has maximum $1$ and minimum $-1\text{.}$ So to find the maximum and minimum of the revenue function $S(x)\text{,}$ we need only to consider the amplitude change and vertical shift. We find,

\begin{equation*} \begin{split} \text{ max } \left(S(x)\right) \amp = \text{ max } \left(\frac{1}{10}\sin\left(\frac{\pi}{2}(x+1)\right)+\frac{1}{2}\right)\\ \amp = \frac{1}{10} \text{ max } \left(\sin\left(\frac{\pi}{2}(x+1)\right)\right) + \frac{1}{2} \\ \amp = \frac{1}{10}(1) + \frac{1}{2} = \frac{3}{5} = 0.6, \end{split} \end{equation*}

or, $600,000. Similarly, we compute \begin{equation*} \text{ min } \left(S(x)\right) = \frac{1}{10}(-1)+\frac{1}{2} = \frac{2}{5} = 0.4\text{,} \end{equation*} or$400,000.

2. Find the values of $x$ where the quarterly revenues are highest and lowest.

$x = 4n+2\text{,}$ $(n=0,1,2,...)\text{.}$
Solution

Now, we need to consider the phase shift given by $f(x)=\frac{\pi}{2}(x+1)\text{.}$ Use the fact that $\sin x$ takes on maximum values at $x = 2\pi m + \frac{\pi}{2}$ ($m$ integer) and minimum values at $x = 2\pi m - \frac{\pi}{2}$ ($m$ integer). Therefore $S(x)$ takes on maxima at

\begin{equation*} \begin{split} f(x) \amp = 2\pi m + \frac{\pi}{2} \\ \frac{\pi}{2}(x+1) \amp = 2\pi m + \frac{\pi}{2}\\ x \amp = 4m. \end{split} \end{equation*}

Note that we require $m=0,1,2,\dots$ since $x$ must be a positive integer. This means that the quarterly revenues are highest when $x=0,4,8,\dots$ . Similarly, we find that the quarterly revenues are lowest when

\begin{equation*} \begin{split} f(x) \amp = 2\pi m - \frac{\pi}{2} \\ \frac{\pi}{2}(x+1) \amp = 2\pi m - \frac{\pi}{2}\\ x \amp = 4m-2, \ \ \ (m = 1,2,3,\dots) \end{split} \end{equation*}

or, when $x=2,6,10,\dots$ . We illustrate with the following sketch:

Compute the following:

1. $\sin^{-1}(\sqrt{3}/2)$

$\pi/3$
Solution

Recall the following:

Therefore,

\begin{equation*} \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}\text{.} \end{equation*}
2. $\cos^{-1}(-\sqrt{2}/2)$

$3\pi/4$
Solution

First, note that $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\text{.}$ From the graphs above, we can see that

\begin{equation*} \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\text{.} \end{equation*}

And so,

\begin{equation*} \cos^{-1}\left(\frac{-\sqrt{2}}{2}\right) = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}\text{.} \end{equation*}

Compute the following:

1. $\sin^{-1}\left(\sin(\pi/4)\right)$

$\pi/4$
Solution
$\displaystyle{\sin^{-1}\left(\sin(\pi/4)\right)= \frac{\pi}{4}}$
2. $\sin^{-1}\left(\sin(17\pi/3)\right)$

$-\pi/3$
Solution
\begin{equation*} \begin{split}\sin^{-1}\left(\sin(17\pi/3)\right)\amp= \sin^{-1}\left(\sin(17\pi/3 - 6\pi)\right) \\ \amp= \sin^{-1}\left(\sin(-\pi/3)\right) = -\frac{\pi}{3}\end{split} \end{equation*}
3. $\cos\left(\cos^{-1}(1/3)\right)$

$1/3$
Solution
$\displaystyle{\cos\left(\cos^{-1}(1/3)\right) = \frac{1}{3}}$
4. $\tan\left(\cos^{-1}(-4/5)\right)$

$-3/4$
Solution

Let $\displaystyle{\theta=\cos^{-1}(-4/5)}\text{.}$ Then $\cos(\theta) = -\dfrac{4}{5}\text{,}$ and we can draw the following diagram:

Thus, $\displaystyle{\tan(\theta) =\theta=\cos^{-1}(-4/5)= -\frac{3}{4}.}$

Rewrite the expression $\tan\left(\cos^{-1}x\right)$ without trigonometric functions. What is the domain of this function?

$\sqrt{1-x^2}/x$ with domain $[-1,0)\cup(0,1]\text{.}$

Let $\theta = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\text{.}$ Evaluate:

1. $\sin \theta$

$\frac{1}{\sqrt{2}}$
2. $\cos \theta$

$\frac{1}{\sqrt{2}}$
3. $\sec \theta$

$\sqrt{2}$
4. $\tan \theta$

$1$
5. $\cot \theta$

$1$
Solution

Let $\displaystyle{\theta = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)}\text{.}$

1. $\sin(\theta) = \dfrac{1}{\sqrt{2}}\text{,}$ and so we can make the following diagram:

Using the above diagram, we see that:

2. $\displaystyle{\cos\theta = \frac{1}{\sqrt{2}}}$

3. $\displaystyle{\sec\theta= \frac{1}{\cos\theta} = \sqrt{2}}$

4. $\displaystyle{\tan\theta= \frac{1}{1} = 1}$

5. $\displaystyle{\cot\theta= \frac{1}{\tan\theta} = 1}$