Question

How do you find the exact value of $\sec \left( \arcsin \left( \dfrac{4}{5} \right) \right)$?

Answer 1

Hint: We start solving the problem by recalling the fact that $\arcsin \left( x \right)={{\sin }^{-1}}\left( x \right)$. We then assign a variable to represent the angle ${{\sin }^{-1}}\left( \dfrac{4}{5} \right)=\alpha $. We then make use of the fact that if $\arcsin \left( x \right)=\theta $, then the values of $\theta $ lies in the interval $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and the values of $\cos \theta $ and $\sec \theta $ is positive in this interval. We then make use of the that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ and make the necessary calculations to get the value of $\cos \left( \arcsin \left( \dfrac{4}{5} \right) \right)$. We then make use of the fact that that $\cos \alpha =\dfrac{1}{\sec \alpha }$ and make necessary calculations to get the required value of $\sec \left( \arcsin \left( \dfrac{4}{5} \right) \right)$.

Complete step by step answer:
According to the problem, we are asked to find the exact value of $\sec \left( \arcsin \left( \dfrac{4}{5} \right) \right)$.
We know that $\arcsin \left( x \right)={{\sin }^{-1}}\left( x \right)$. So, we get $\arcsin \left( \dfrac{4}{5} \right)={{\sin }^{-1}}\left( \dfrac{4}{5} \right)$.
Let us assume ${{\sin }^{-1}}\left( \dfrac{4}{5} \right)=\alpha \Leftrightarrow \sin \alpha =\dfrac{4}{5}$ ---(1).
So, we get \[\sec \left( \arcsin \left( \dfrac{4}{5} \right) \right)=\sec \left( \alpha \right)\] ---(2).
Let us recall the properties of $\arcsin \left( x \right)$. We know that if $\arcsin \left( x \right)=\theta $, then the values of $\theta $ lies in the interval $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. We know that the values of $\cos \theta $ and $\sec \theta $ is positive in this interval.
We know that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$.
From equation (1), we have ${{\left( \dfrac{4}{5} \right)}^{2}}+{{\cos }^{2}}\alpha =1$.
$\Rightarrow \dfrac{16}{25}+{{\cos }^{2}}\alpha =1$.
$\Rightarrow {{\cos }^{2}}\alpha =1-\dfrac{16}{25}$.
$\Rightarrow {{\cos }^{2}}\alpha =\dfrac{25-16}{25}$.
$\Rightarrow {{\cos }^{2}}\alpha =\dfrac{9}{25}$.
$\Rightarrow \cos \alpha =\sqrt{\dfrac{9}{25}}$.
We need to consider the positive square root as we know that the value of cosine function must be positive.
$\Rightarrow \cos \alpha =\dfrac{3}{5}$ ---(3).
We know that $\cos \alpha =\dfrac{1}{\sec \alpha }$. Let us use this result in equation (3).
$\Rightarrow \dfrac{1}{\sec \alpha }=\dfrac{3}{5}$.
$\Rightarrow \sec \alpha =\dfrac{5}{3}$.
From equation (2), we get \[\sec \left( \arcsin \left( \dfrac{4}{5} \right) \right)=\dfrac{5}{3}\].

$\therefore $ We have found the exact value of $\sec \left( \arcsin \left( \dfrac{4}{5} \right) \right)$ as $\dfrac{5}{3}$.

Note:Here we have assumed that the value of ${{\sin }^{-1}}\left( \dfrac{4}{5} \right)$ lies in the principal range of $\arcsin x$ otherwise the answer would have been different. We should not take the negative square root while finding the value of $\cos \alpha $ which is the common mistake done by students. Whenever we get this type of problem, we first assume a variable for the inverse function given in the problem. Similarly, we can expect problems to find the value of $\tan \left( \arccos \left( \dfrac{-4}{5} \right) \right)$.