Question

How do you evaluate $\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right)$?

Answer 1

Hint: In this problem we need to calculate the value of $\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right)$. In the given function we can observe the $\text{arcsin}$ function which is nothing but the inverse trigonometric function of $\sin $, mathematically it is ${{\sin }^{-1}}$ function. So, we will assume the ${{\sin }^{-1}}$ function to be a variable, say $y$. Now we will apply the trigonometric function $\sin $ on both sides of the equation and simplifies the equation to get the value of $\sin y$. After calculating the value of $\sin y$. We will use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to find the value of $\cos y$ which is nothing but our required value.

Complete step by step answer:
Given that, $\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right)$.
In the above equation $\text{arcsin}$ function is nothing but the ${{\sin }^{-1}}$ function. So, we are replacing it in the above equation, then we will get
$\cos \left( \text{arcsin}\left( \dfrac{3}{5} \right) \right)=\cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)$
Considering the function ${{\sin }^{-1}}$ individually, and assuming it to a variable say $y$, then we will get
$y={{\sin }^{-1}}\left( \dfrac{3}{5} \right).....\left( \text{i} \right)$
Applying the trigonometric function $\sin $ on both sides of the above equation, then we will get
$\Rightarrow \sin \left( y \right)=\sin \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)$
We know that the value of $\sin \left( {{\sin }^{-1}}\left( x \right) \right)={{\sin }^{-1}}\left( \sin \left( x \right) \right)=x$, then we will get
$\Rightarrow \sin y=\dfrac{3}{5}$
From the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we can write the value of $\cos y$ as
$\cos y=\sqrt{1-{{\sin }^{2}}y}$
Substituting the value of $\sin y=\dfrac{3}{5}$ in the above equation, then we will get
$\begin{align}
  & \cos y=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \\
 & \Rightarrow \cos y=\sqrt{1-\dfrac{9}{25}} \\
\end{align}$
Simplifying the above equation by taking LCM, then we will have
$\begin{align}
  & \Rightarrow \cos y=\sqrt{\dfrac{25\times 1-9}{25}} \\
 & \Rightarrow \cos y=\sqrt{\dfrac{16}{25}} \\
\end{align}$
We can write $16={{4}^{2}}$, $25={{5}^{2}}$. Now the above equation is modified as
$\begin{align}
  & \Rightarrow \cos y=\sqrt{{{\left( \dfrac{4}{5} \right)}^{2}}} \\
 & \Rightarrow \cos y=\dfrac{4}{5} \\
\end{align}$
From equation $\left( \text{i} \right)$, substituting the value of $y$ in the above equation, then we will get

$\therefore \cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)=\dfrac{4}{5}$

Note: In this problem we have the used the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ after calculating the value of $\sin y$. We can also follow another method for finding the solution after calculating $\sin y$, which is by using the basic definition of trigonometric ratio $\sin x$ we will compare it with the value of $\sin y$. Now we will construct a right-angled triangle with the data we have from the above comparison and try to calculate the remaining data by using the Pythagoras theorem. After having all the data of the triangle, we can use definitions of basic trigonometric ratios and calculate whatever ratio we want.