Question

For the principal values, evaluate each of the following: \[{{\tan }^{-1}}\left( 2\sin \left( 4{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} \right) \right)\]

Answer 1

Hint: To solve the question given above, we will first find out the value of \[4{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\]. We will assume that it is the value of x. Then we will find the value of \[2\sin x\]. We will assume that it is value of y. Then finally, we will calculate the value of \[{{\tan }^{-1}}y\]. We will assume that the value of the whole term evaluates to be z.

Complete step-by-step solution:
To start with, we will assume that the value of the term given in the question will be z. Thus, we will get:
\[z={{\tan }^{-1}}\left( 2\sin \left( 4{{\cos }^{-1}}\dfrac{\sqrt{3}}{2} \right) \right)\] ---- (1)
Now, we will assume that the value of \[4{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] is x. Thus, we will get the following equation:
\[z={{\tan }^{-1}}\left( 2\sin x \right)\] ----- (2)
Now, we will assume that the value of \[2\sin x\] is y. Thus, we will get the following equation:
\[z={{\tan }^{-1}}\left( y \right)\] --------- (3)
Now, we will calculate the value of x. For this, we will have to find the value of \[{{\cos }^{-1}}\dfrac{\sqrt{3}}{2}\]. We know that:
\[\cos \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}\]
Now, we will take \[{{\cos }^{-1}}\] on both sides. Thus, we will get following equations:
\[{{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{3} \right) \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\]
Now, we will apply the following identity:
\[{{\cos }^{-1}}\left( \cos x \right)=x\] where $x \in [0, \pi]$
Thus, we will get:
\[{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{3}\]
Now the value of \[x=4{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\]. Thus, we will get:
\[\Rightarrow x=\dfrac{4\pi }{3}\]
Now, we have to find the value of y. For this, we must know the value of \[\sin x\].
\[\begin{align}
  & \sin x=\sin \left( \dfrac{4\pi }{3} \right) \\
 & \sin x=\sin \left( \dfrac{3\pi +\pi }{3} \right) \\
 & \Rightarrow \sin x=\sin \left( \pi +\dfrac{\pi }{3} \right) \\
\end{align}\]
Now, we will use the following identity: \[\sin \left( \pi +x \right)=-\sin x\].
Thus, we will get the following equation:
\[\Rightarrow \sin x=-\sin \left( \dfrac{\pi }{3} \right)\]
Now, we know that the value of \[\sin \left( \dfrac{\pi }{3} \right)\] is \[\dfrac{\sqrt{3}}{2}\]. So, we will get following equation:
\[\Rightarrow \sin x=-\dfrac{\sqrt{3}}{2}\]
Now, the value of \[y=2\sin x\].
\[\begin{align}
  & \Rightarrow y=2\left( -\dfrac{\sqrt{3}}{2} \right) \\
 & \Rightarrow y=-\sqrt{3} \\
\end{align}\]
Now, we will calculate the value of z. We know that, from equation (3):
\[z={{\tan }^{-1}}y\]
\[z={{\tan }^{-1}}\left( -\sqrt{3} \right)\]
We can write as, \[{{\tan }^{-1}}\left( -\sqrt{3} \right)\] as \[-{{\tan }^{-1}}\left( \sqrt{3} \right)\] because \[{{\tan }^{-1}}\left( x \right)\] is an odd function. So we will get:
\[\Rightarrow z=-{{\tan }^{-1}}\sqrt{3}\] --------- (4)
Now, we know that \[\tan \dfrac{\pi }{3}=\sqrt{3}\]. We will take \[{{\tan }^{-1}}\] on both sides. Thus we will get:
\[{{\tan }^{-1}}\left( \tan \dfrac{\pi }{3} \right)={{\tan }^{-1}}\left( \sqrt{3} \right)\]
Now, we will use the identity: \[{{\tan }^{-1}}\left( \tan x \right)=x\] where $x \in (\dfrac{-\pi}{2}, \dfrac{\pi}{2})$. Thus, we will get:
\[\dfrac{\pi }{3}={{\tan }^{-1}}\left( \sqrt{3} \right)\]
\[\Rightarrow {{\tan }^{-1}}\left( \sqrt{3} \right)=\dfrac{\pi }{3}\] ------- (5)
From (4) and (5), we have:
\[z=-\dfrac{\pi }{3}\] --------- (6)
From (1) and (6), we have:
\[{{\tan }^{-1}}\left( 2\sin \left( 4{{\cos }^{-1}}\dfrac{\pi }{3} \right) \right)=-\dfrac{\pi }{3}\]

Note: While solving the question, we have used the following identity: \[{{\cos }^{-1}}\left( \cos x \right)=x\]. This identity is not valid everywhere. This is valid only when x lies between 0 and $\pi$. If x does not lie in this interval, then we have to make the necessary changes in identity.