Question

How do you use an inverse trigonometric function to solve equations?

Answer 1

First regardless how you are used to dealing with exponentiation we tend to denote an inverse trigonometric function with an “Exponent” of -1.
For example:
The inverse of cosine is denoted as ${\cos ^{ - 1}}\left( x \right)$
In this case we have to note that $- 1$ is not an exponent ${\cos ^{ - 1}}x \ne \dfrac{1}{{\cos x}}$
In trig function $- 1$ looks like an exponent but it isn't. It is simply a notation that we use to denote that we are dealing with an inverse trig function.
If I had really wanted exponentiation to denote $1$ over cosine would use the following.
${\left( {\cos \left( x \right)} \right)^{ - 1}} = \dfrac{1}{{\cos \left( x \right)}}$
There are many other inverse functions but you will mainly see these three functions in calculus.
To evaluate inverse trigonometric function. Remember that the following statements are equivalent.
$\theta = {\cos ^{ - 1}}\left( x \right) \Rightarrow x = \cos \left( \theta \right)$
$\theta = {\sin ^{ - 1}}\left( x \right) \Rightarrow x = \sin \left( \theta \right)$
$\theta = {\tan ^{ - 1}}\left( x \right) \Rightarrow x = \tan \left( \theta \right)$

Complete step by step solution:
Let’s understand deeply about inverse trigonometric functions.
Let’s take an example find the value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
Let $t = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
$\cos \left( t \right) = \dfrac{{\sqrt 3 }}{2}$
In other words we asked what angle $t$ do we need to plug into cosine to get the value $\dfrac{{\sqrt 3 }}{2}?$ This is essentially what we are asking here when we are asked to compute the inverse trigonometric function.
${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
In this problem we are solving an equation which has an intuit number as a solution. These were
$\dfrac{\pi }{6} + 2\pi n,n = 0, \pm 1, \pm 2, \pm 3....$
$\dfrac{{11\pi }}{6} + 2\pi n,n = 0, \pm 1, \pm 2, \pm 3,....$
In this case, at a trigonometric function we need a single value answer and not the infinite possible answer. So, to make sure we get a single value out of the inverse trigonometric cosine function we use the following restriction on inverse cosine.
$\theta = {\cos ^{ - 1}}\left( x \right)$ $- 1 \leqslant x \leqslant 1$ and $0 \leqslant \theta \leqslant \pi$
The restriction on $\theta$ guarantees that we will only get a single value angle and we can’t get values of $x$ out of cosine that are larger than $1$ or smaller than $- 1$
So, using the restriction.
${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{6}$

Note: We put a restriction on angle i.e. $0 \leqslant \theta \leqslant \pi$. Than value of $n$ and check the correct value which should not exceed $\pi$ in
$\dfrac{\pi }{6} + 2\pi n$