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

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 $\sin x$.

& \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{aligned}$$ Now, we will use the following identity: $$\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{aligned} & \Rightarrow y=2\left( -\dfrac{\sqrt{3}}{2} \right) \\\ & \Rightarrow y=-\sqrt{3} \\\ \end{aligned}$$ Now, we will calculate the value of z. We know that, from equation (3): $$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.