8.5 More (and Inverse) Trig Functions

Learning Objectives

By the end of this section, you will be able to:

• recognize and graph transformations of sine and cosine
• understand and recognize the inverse trigonometric functions and their domains
• evaluate inverse trig functions

Transformations of Trig Functions

Vocab-wise, we mentioned the period of a trig function last section: the amount of time it takes for it to complete one full cycle of behavior and start over. To help you out in future science classes, I want to make sure we cover a couple more terms. The amplitude of a sine or cosine wave is the maximum distance between the graph of the function and the horizontal axis; i.e. the max height (and max dip). If you perform a horizontal shift on a sine or cosine wave, you might hear the term phase shift. We'll start with amplitude:

The amplitude of \( f(x) = \sin x\) is just \(1\), since sine only goes as high as \(1\) and as low as \(-1\). The blue graph is \( g(x) = 2\sin x\), and we recall that multiplying a whole function by a constant results in a vertical stretch, which has the effect of increasing the amplitude to \(2\). Both of these functions complete a full cycle between \(x = 0\) and \(x = 2\pi\), and then repeat themselves forever, so their period is \(2\pi\). Now consider the next example:

This time, \(f(x) = \sin x\) has had its input replaced by \( \left( x - \frac{\pi}{4} \right) \), which we recall causes a horizontal shift to the right. Last transformation type:

This last picture shows the difference between \( f(x) = \sin x\) and \(g(x) = \sin (2x)\). Multiplying by \(2\) inside the input causes a horizontal shrink, and we see that all of the "milestones" on the green graph (max values, min values, hitting the \(x\)-axis...) are happening "sooner" on the blue graph. Instead of peaking at \(x = \frac{\pi}{2}\), the squished version peaks sooner at \(x = \frac{\pi}{4}\), etc. This horizontal squish also affects the period, because the blue graph finishes a cycle twice as fast, making his period just \(\pi\).

For a function of the form \(f(x) = a \sin (x + c) \) or \(a \cos (x + c)\),

• \(|a|\) is the amplitude
• \(c\) gives the phase shift (to the left if \(+\), to the right if \(-\))
• the period is \(2\pi\)

For a function of the form \( f(x) = a \sin (bx) \) or \(a \cos (bx)\),

• \(|a|\) is the amplitude
• \(b\) causes a horizontal shrink if \(b>1\), and a horizontal stretch if \(0<b<1\)
• the period is \( \frac{2\pi}{b}\)

To sketch the graph of a transformed trig function,

1. Sketch the OG trig function first with its important signpost points (max values, min values, roots, basically the info you get from the cardinal points on the unit circle). This will help you imagine what the transformed trig function should look like based on the transformations that happened. 
2. Determine the transformations that have occurred based on the function's definition. Analyze how they change the signpost points. This can be done using common sense or with a from-the-ground-up approach, so read the following examples carefully to see strategies.
3. Plot the new signpost points and smoothly connect the dots.

Example \(\PageIndex{1}\)

Sketch the graph of \( f(x) = 2 \cos (x + \pi) \).

Solution

First, we sketch plain ol' \(\cos x\) with his important points:

Next, we identify that the multiplication by \(2\) should cause a vertical shift, stretching the amplitude to \(2\). The \( (x+\pi)\) will cause a horizontal shift to the left by \(\pi\). Here is how that will affect the signpost points, performing both transformations one after another, moving from left column to right:

Original	After Vertical Stretch (multiply \(y\)-coordinate only)	After Horizontal Shift (subtract \(\pi\) to slide \(x\)-coordinates to the left)
\( (0,1)\)	\( (0,\textcolor{red}{2})\)	\( (\textcolor{red}{-\pi},2)\)
\( (\pi/2, 0)\)	\( (\pi/2,\textcolor{red}{0})\)	\( (\textcolor{red}{-\pi/2}, 0)\)
\( (\pi, -1)\)	\( (\pi, \textcolor{red}{-2})\)	\( (\textcolor{red}{0}, -2)\)
\( (3\pi/2,0) \)	\( (3\pi/2, \textcolor{red}{0})\)	\( (\textcolor{red}{\pi/2}, 0) \)
\( (2\pi, 1) \)	\( (2\pi,\textcolor{red}{2})\)	\( (\textcolor{red}{\pi},2) \)

Now, we plot the new (third column) points and connect the dots. The original \( \cos x\) is shown in dotted blue if you want to compare.

Exercise \(\PageIndex{1}\)

Sketch the graph of \(f(x) = \frac{1}{2} \sin \left( x + \frac{\pi}{2} \right)\).

Answer

Original	After Vertical Shrink	After Phase Shift
\( (0,0)\)	\( (0,0)\)	\( (-\pi/2, 0)\)
\( (\pi/2,1)\)	\( (\pi/2, 1/2) \)	\( (0, 1/2)\)
\( (\pi,0)\)	\( (\pi,0)\)	\( (\pi/2, 0)\)
\( (3\pi/2, -1)\)	\( (3\pi/2, -1/2) \)	\( (\pi, -1/2)\)
\( (2\pi,0)\)	\( (2\pi,0)\)	\( (3\pi/2, 0)\)

Example \(\PageIndex{2}\)

Sketch the graph of \( f(x) = -3 \sin (2x) \).

Solution

We see that \(-3\) multiplication out front and recognize two things: we will have a stretched amplitude of \(3\), and our graph will have been reflected across the \(x\)-axis. We also see \(2x\) in the input. Basically, \(\sin (2x)\) does everything \( \sin x\) does, but faster, at smaller \(x\)-values. If \(\sin x\) is 1 at \(\pi/2\), then \( \sin (2x)\) will reach \(1\) at \(\pi/4\) instead. I'm going to break this down into a few steps over some columns: the first column will be the \(x\)-coordinates for some sample points. The second column will be \(2x\), which then is the input for sine in the third column. Finally, we multiply by \(-3\) to get our final \(y\)-values.

\(x\)-value	\(2x\)	\( \sin(2x)\)	\( -3 \sin(2x) \)	Points
\( 0\)	\(0\)	\(0\)	\(0\)	\( (0,0) \)
\( \pi/4\)	\( \pi/2\)	\(1\)	\(-3\)	\( (\pi/4, -3)\)
\( \pi/2\)	\( \pi\)	\(0\)	\(0\)	\( (\pi/2, 0)\)
\( 3\pi/4\)	\( 3\pi/2\)	\(-1\)	\(3\)	\( (3\pi/4, 3)\)
\( \pi\)	\(2\pi\)	\(0\)	\(0\)	\( (\pi, 0)\)

One more note: I chose my sample points to be small increments in fourths, knowing that when they're multiplied by \(2\) in the second column, they will become the easy values \(0, \pi/2, \pi,\) etc. Now I plot the points and connect the dots, extending the periodic behavior.

Exercise \(\PageIndex{2}\)

Sketch the graph of \( f(x) = 2\cos \left( \frac{1}{2} x \right) \).

Answer

Hint: choose sample \(x\)-values such that they can easily be divided by 2 and stay nice!

\(x\)-value	\(\frac{1}{2}x\)	\( \cos \left( \frac{1}{2} x \right) \)	\( 2\cos \left( \frac{1}{2} x \right) \)	Points
\(0\)	\( 0\)	\(1\)	\(2\)	\( (0,2)\)
\(\pi\)	\( \pi/2\)	\(0\)	\(0\)	\( (\pi,0)\)
\(2\pi \)	\( \pi\)	\(-1\)	\(-2\)	\( (2\pi, -2) \)
\( 3\pi \)	\( 3\pi/2\)	\(0\)	\(0\)	\( (3\pi, 0)\)
\(4\pi \)	\( 2\pi\)	\(1\)	\(2\)	\( (4\pi,2)\)

Inverse Trig Functions

Inverse functions "undo" each other, as we recall. Betcha can't guess what inverse trig functions do.

"Hold the phone!" you say? "Not so fast... Only one-to-one functions can have inverses!" You're completely correct. And yes, sine, cosine, and tangent all fail the Horizontal Line Test, so I guess we can all close our books and go home!

Wait, come back, I have a solution. We're going to zoom in on our trig functions and only consider them on a particular domain. For example, if I only consider \( \sin x\) for \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\) (the "right half" of the unit circle), then he passes the HLT with flying colors.

It's perfectly reasonable to define an inverse function for this little piece. Recalling that inverse functions look like reflections across the diagonal line \( y = x \), we expect it to look like this:

This new red function is called arcsine. Whereas we feed sine an angle and it returns a number between \(-1\) and \(1\), we turn around and feed arcsine a number between \(-1\) and \(1\), and it returns an angle between \(-\frac{\pi}{2}\) and \( \frac{\pi}{2}\). As always with inverse functions, their domains and ranges are swapped!

For cosine and tangent, we can also restrict their domains to a particular piece (called a "branch") so that they pass the HLT, and define their inverse functions, which are called arccosine and arctangent. I will always use this notation:

\[ \arcsin x \quad \quad \arccos x \quad \quad \arctan x \notag \]

but you will also come across the notation style \( \sin^{-1} x, \cos^{-1} x, \tan^{-1} x \), because of how we've written \(f^{-1}\) to mean the inverse of \(f\). In particular, you'll see this on calculators because they want the buttons to be small and they can't fit "arc" on them... However, I find this notation morally and pragmatically reprehensible, because students will often interpret the \(^{-1}\) to mean reciprocal, like \(4^{-1} = \frac{1}{4}\). So I'm a big "arc" apologist. Here's the gist on the Big Three.

Function	Observations
	Domain: arcsine only takes inputs \( -1 \leq x \leq 1\), because those are the only outputs of sine. In interval notation, dom: \( [-1, 1]\). Range: arcsine only outputs values from\( \frac{-\pi}{2}\) to \( \frac{\pi}{2} \), because that was the restricted domain we chose for sine. You can think of those angles as the "right half" of the unit circle. In interval notation, range: \( [-\pi/2, \pi/2]\). Arcsine passes through \( (0,0)\) because \( \sin(\textcolor{red}{0}) = \textcolor{blue}{0}\), so we must have \( \arcsin(\textcolor{blue}{0}) = \textcolor{red}{0} \). Similarly, it has signpost endpoints \( (-1, -\pi/2)\) and \( (1, \pi/2) \).
	Domain: arccosine takes inputs \( -1 \leq x \leq 1\) for the same reason as arcsine. In interval notation, dom: \( [-1, 1]\). Range: arccosine only spits out values from \( 0\) to \( \pi\), because that's the restricted domain we use for cosine (by tradition). You can remember those angles as the "top half" of the unit circle. In interval notation, range: \( [0, \pi]\). Arccosine's signpost points are the endpoints \( (-1,\pi)\) and \( (1,0) \), and the \(y\)-intercept \( (0, \pi/2) \).
	Domain: Arctangent takes all real numbers as inputs, because that's the range of tangent. In interval notation, dom: \( (-\infty, \infty)\). Range: Arctangent spits out values strictly between \( -\pi/2\) and \( \pi/2 \), because that's the conventional restricted domain of tangent. In interval notation, range: \( (-\pi/2, \pi/2)\). Going off to the right forever, arctangent approaches the horizontal asymptote \( y = \pi/2\) from below. Going off to the left, it approaches \(y = - \pi/2\) from above. When sketching arctangent, you should plot the easy points \( (0,0), (-1, -\pi/4),\) and \( (1, \pi/4) \), drop in your horizontal asymptotes, and then smoothly swoop in through the points.

There are, of course, inverse functions for the reciprocal trig functions; these are arccosecant (\( \operatorname{arccsc} x\)), arcsecant (\( \operatorname{arcsec} x\)), and arccotangent (\(\operatorname{arccot} x\)), but for the life of me I can't remember needing to have them on instant recall in my other courses. I'll guide you through exploring those a bit in the exercises, and assume that if you need them later in life, you can Google them along the way.

Evaluating inverse trig functions is a total scam, because all you do is translate back into regular trig world, sketch a unit circle, and use your brain.

Example \(\PageIndex{3}\)

Evaluate.

1. \(\arcsin(-1) \)
2. \( \arccos\left( \frac{\sqrt{2}}{2} \right) \)
3. \( \arctan (\sqrt{3}) \)

Solution

1. We want to find \( \arcsin(-1) = ?\), and he spits out angles, so I want to find the \( \theta\) such that \( \sin \theta = -1\). Well, where is sine \(-1\)? Remember your unit circle... At \( \frac{3\pi}{2}\), with coordinates \( (0,-1)\), sine is \(-1\). BUT WAIT, that's not in our allowed range of angles for arcsine. Simply recall that \( \frac{3\pi}{2}\) is coterminal with the negative angle \( -\frac{\pi}{2}\). That's our final answer.

2. Again, translate to regular trig. We are looking for the angle \(\theta\) where \( \cos \theta = \frac{\sqrt{2}}{2} \).That's just \( \frac{1}{\sqrt{2}}\) after rationalizing the denominator. You will see people write it this way OFTEN. So anyways, on the unit circle, cosine has this value at two angles: \( \frac{\pi}{4}\) and \( \frac{7\pi}{4}\). We only report the first one as our answer, because the other option is not in our allowable range!

3. We are looking for \(\theta\) such that \( \tan \theta = \sqrt{3}\). In other words, we need \(\dfrac{ \sin \theta}{\cos \theta} = \sqrt{3} \). With the unit circle, we notice that at \(\frac{\pi}{3}\), we have coordinates \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \). Remember that those are (cosine, sine) values, so if you divide them in the correct order, the fraction simplifies to \( \sqrt{3}\). This angle is in our allowed range, so we report back \( \frac{\pi}{3}\).

Exercise \(\PageIndex{3}\)

Evaluate.

1. \( \arccos(1)\)
2. \( \arcsin \left(\frac{1}{2} \right) \)
3. \( \arctan (-1) \)

Answer

1. \(0\)
2. \( \pi/6\)
3. \( -\pi/4\)

Okay, you can go do a lot more practice with those in the exercises!