Question

Find the values of each of the following:
${{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]$

Answer 1

Hint: We will apply the formula $\sin \left( x \right)=\sin \left( y \right)$ then $x=n\pi +{{\left( -1 \right)}^{n}}y$. Here n belongs to the integers. We will use this formula in order to find the value of the angle of sine.

Complete step-by-step answer:
First we will consider ${{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]...(i)$. We will substitute the value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=a$. By placing the inverse sin operation to the right side of the equal sign then we will have $\sin \left( a \right)=\dfrac{1}{2}$.
As we know that the value of $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$. Therefore, we have that $\sin \left( a \right)=\sin \left( \dfrac{\pi }{6} \right)$. The sine is positive in first and second quadrants. So, we will consider the sine in the first quadrant only due to the range of inverse sine which is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. Therefore, we have that $a=\dfrac{\pi }{6}$. This is by the formula if we have $\sin \left( x \right)=\sin \left( y \right)$ then $x=n\pi +{{\left( -1 \right)}^{n}}y$. Here n belongs to the integers. As we already have ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=a$ so, we can write this expression as ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}$. Now, we will substitute this value in the equation (i). Thus, we have
$\begin{align}
  & {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]={{\tan }^{-1}}\left[ 2\cos \left( 2\times \dfrac{\pi }{6} \right) \right] \\
 & \Rightarrow {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]={{\tan }^{-1}}\left[ 2\cos \left( \dfrac{\pi }{3} \right) \right]...(ii) \\
\end{align}$
Now as we know that the value of $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$. Therefore, after substituting the value in equation (ii) we will have
$\begin{align}
  & {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]={{\tan }^{-1}}\left[ 2\cos \left( \dfrac{\pi }{3} \right) \right] \\
 & \Rightarrow {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]={{\tan }^{-1}}\left[ 2\times \dfrac{1}{2} \right] \\
 & \Rightarrow {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]={{\tan }^{-1}}\left( 1 \right) \\
\end{align}$
Now we will find the value of ${{\tan }^{-1}}\left( 1 \right)$. Since, the value of $\tan \left( \dfrac{\pi }{4} \right)=1$ so we will get
$\begin{align}
  & {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]={{\tan }^{-1}}\left( 1 \right) \\
 & \Rightarrow {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4} \right) \right) \\
\end{align}$
Now we will apply the formula of ${{\tan }^{-1}}\left( \tan \left( x \right) \right)=x$ in the above expression. Therefore, we have
 $\Rightarrow {{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]=\dfrac{\pi }{4}$
And this is our required answer.
Hence, the value of the expression ${{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]=\dfrac{\pi }{4}$.

Note: We could have used an alternate method of solving ${{\tan }^{-1}}\left[ 2\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) \right]$.We will substitute the value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=a$. By placing the inverse sin operation to the right side of the equal sign then we will have $\sin \left( a \right)=\dfrac{1}{2}$. And here we could have used the trigonometric identity which is given by ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ or, ${{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x$ . Therefore, we get
$\begin{align}
  & {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x \\
 & \Rightarrow \dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right) \right) \\
\end{align}$
This is because the value of $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$. Therefore we could have got the desired result by this method also.