7.02 Inverse trigonometric functions

Introduction to inverse trigonometric functions

Finding an inverse for $\sin x$sinx

Consider the equation:

$\sin x=\frac{1}{\sqrt{2}}$sinx=1√2​

We would like to find the solutions to this equation, shown as the points of intersection on the graph below. To begin with we are going to determine an inverse function for sine, which will help us find one of these solutions. We will then look at how to recover the other solutions in a later lesson.

 

The graph of an inverse function can be found by reflecting the original function about the line $y=x$y=x. The graph below shows that reflection applied to the graph of $y=\sin x$y=sinx.

 

Notice, however, that this graph has multiple $y$y-values for a given $x$x-value, so it is not a function.

 

In order to find an inverse function for $\sin x$sinx we will need to restrict the domain of the function to a portion that meets the following criteria:

1. It only has one $y$y-value for each $x$x-value (it is one-to-one).
2. It covers the full range of values of $\sin x$sinx.

There are infinitely many ways we could restrict the domain that would satisfy these criteria, but the domain that is typically chosen by mathematicians to work with is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2​,π2​]. When just this part of the curve is reflected about $y=x$y=x we obtain the inverse function for $\sin x$sinx, which we represent as $\sin^{-1}\left(x\right)$sin−1(x) or $\arcsin(x)$arcsin(x).

We can now rewrite our original equation using the inverse sine function:

$\sin x$sinx	$=$=	$\frac{1}{\sqrt{2}}$1√2​
$x$x	$=$=	$\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$sin−1(1√2​)
$x$x	$=$=	$\frac{\pi}{4}$π4​

Notice that because we needed to restrict the domain of $\sin x$sinx to obtain an inverse function, we have only found the one solution between $-\frac{\pi}{2}$−π2​ and $\frac{\pi}{2}$π2​. We will look at how to use this single solution to find the other solutions to the equation later.

 

Finding an inverse for $\cos x$cosx and $\tan x$tanx

We can find inverse functions for $\cos x$cosx and $\tan x$tanx in a similar way; we first restrict their domains using the above criteria, then reflect the remaining graph about the line $y=x$y=x.

The domain of $\cos x$cosx is restricted to $\left[0,\pi\right]$[0,π] to produce the inverse function, which we represent as $\cos^{-1}\left(x\right)$cos−1(x) or $\arccos(x)$arccos(x).

 

The domain of $\tan x$tanx is restricted to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2​,π2​) to produce the inverse function, which we represent as $\tan^{-1}\left(x\right)$tan−1(x) or $\arctan(x)$arctan(x).

Practice question

Question 1

 

Evaluating inverse trigonometric functions

The inverse trigonometric functions, $\arcsin$arcsin, $\arccos$arccos and $\arctan$arctan are notated on calculators, somewhat confusingly, as $\sin^{-1}$sin−1, $\cos^{-1}$cos−1 and $\tan^{-1}$tan−1 respectively.

It is important to realize that the notation refers to the function inverses and not to the multiplicative inverses or reciprocal functions.

Functions and their inverses have the property that, for a function $f$f, it is always true that $f\left(f^{-1}(x)\right)=x$f(f−1(x))=x and also, $f^{-1}\left(f(x)\right)=x$f−1(f(x))=x.

In the case of the trigonometric functions, this means 

$\sin\left(\arcsin x\right)$sin(arcsinx)	$=$=	$x$x
$\arcsin\left(\sin x\right)$arcsin(sinx)	$=$=	$x$x
$\cos\left(\arccos x\right)$cos(arccosx)	$=$=	$x$x
$\arccos\left(\cos x\right)$arccos(cosx)	$=$=	$x$x
$\tan\left(\arctan x\right)$tan(arctanx)	$=$=	$x$x
$\arctan\left(\tan x\right)$arctan(tanx)	$=$=	$x$x

 

Given a number that is in the range of either the sine, cosine or tangent function, we may ask the question, 'What is the number or angle that has this sine, cosine or tangent'. This is the situation in which we use the inverse trigonometric functions.

For example, the number $0.44352$0.44352 is in the range of the cosine function. That is, there is a number $\theta$θ such that $\cos\theta=0.44352$cosθ=0.44352.

We may wish to find the $\theta$θ whose cosine is $0.44352$0.44352 and for this, we apply the inverse cosine function, $\cos^{-1}(0.44352)$cos−1(0.44352) and obtain the result $\theta=63.67^\circ$θ=63.67°.

 

Worked examples

Question 2

Determine: We are told that the cosine of an angle is $\frac{\sqrt{3}}{2}$√32​. Find an angle between $0$0 and $2\pi$2π that has this cosine.

Think: We need to evaluate $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$cos−1(√32​).

Do: This can be done by calculator, but a more elegant solution is to recognize the quantity $\frac{\sqrt{3}}{2}$√32​ as an exact value of the cosine function.

Either way, we obtain the solution  $\frac{\pi}{6}$π6​ .

 

Question 3

Evaluate $\cos\left(\sin^{-1}\left(\frac{13}{85}\right)\right)$cos(sin−1(1385​)).

This can be done in two ways.

Strategy 1 - Think: First, if the sine of an angle is $\frac{13}{85}$1385​, then the length of the opposite side would be $13$13 and the length of the hypotenuse would be $85$85.

Strategy 1 - Do: Using the Pythagorean Theorem, we find that this makes the length of the adjacent side $84$84. The cosine of the angle must be $\frac{84}{85}$8485​. This can be seen in the diagram below.

So, we get

$\cos\left(\sin^{-1}\left(\frac{13}{85}\right)\right)$cos(sin−1(1385​))	$=$=	$\cos\left(\cos^{-1}\left(\frac{84}{85}\right)\right)$cos(cos−1(8485​))	Using the triangle above
	$=$=	$\frac{84}{85}$8485​	Definition of the inverse

 

Strategy 2 - Think: A second way to use our calculator to evaluate the expression. We can do this in one or two steps. 

Strategy 2 - Do: Doing it in two steps on the calculator

$\cos\left(\sin^{-1}\left(\frac{13}{85}\right)\right)$cos(sin−1(1385​))	$=$=	$\cos(8.79741071\ldots^\circ)$cos(8.79741071…°)
	$=$=	$0.9882352941$0.9882352941...
	$=$=	$\frac{84}{85}$8485​

 

Find the angle when given a trigonometric measure

As with every topic in mathematics, there is a conceptual side (what you need to know and understand) and a practical side (what you need to do and answer). To calculate values involving trigonometric expressions, it will often be easiest to use a scientific calculator.

Don't forget the three trigonometric ratios!

 

 

Worked examples

Question 4

If $\sin\theta=0.65$sinθ=0.65, find $\theta$θ to the nearest degree.

Think: This question is asking us what the angle ($\theta$θ) is, if the ratio of the opposite side and hypotenuse is $0.65$0.65. To answer this question, we can use the inverse sine button on a calculator. It will probably look like $\sin^{-1}$sin−1, and may involving pressing 'shift' or '2nd F'.

Do:

$\sin\theta$sinθ	$=$=	$0.65$0.65
$\theta$θ	$=$=	$\sin^{-1}\left(0.65\right)$sin−1(0.65)	(Take the inverse sine)
$\theta$θ	$=$=	$40.54160187$40.54160187$\ldots$…	(Evaluate with a calculator)
$\theta$θ	$=$=	$41^\circ$41°	(Round to the nearest degree)

Practice questions

Question 5

QUESTION 6

Question 7

Question 8

QUESTION 9