Question

For the principal value, evaluate the following
$${\sin }^{-1}[\cos \left(2{\mathrm{cosec}}^{-1}(-2)\right)]$$

Answer 1

Hint:First of all, use $${\mathrm{cosec}}^{-1}(-x)=-{\mathrm{cosec}}^{-1}x$$ and then use a trigonometric table to find the value of $${\mathrm{cosec}}^{-1}\left(2\right)$$. Now use $$\cos (-\theta )=\cos \theta$$ and again use the table to find the value of $$\cos {\displaystyle \frac{\pi }{3}}$$. Find the angle at which $$\sin \theta ={\displaystyle \frac{1}{2}}$$ from the table or the value of $${\sin }^{-1}\left({\displaystyle \frac{1}{2}}\right)$$ to get the required answer.

Complete step-by-step answer:
In this question, we have to find the principal value of $${\sin }^{-1}[\cos \left(2{\mathrm{cosec}}^{-1}(-2)\right)]$$.
First of all, let us consider the expression given in the question,
$$E={\sin }^{-1}[\cos \left(2{\mathrm{cosec}}^{-1}(-2)\right)]$$
We know that, $${\mathrm{cosec}}^{-1}(-x)=-{\mathrm{cosec}}^{-1}x$$. By using this in the above expression, we get,
$$E={\sin }^{-1}[\cos (-2{\mathrm{cosec}}^{-1}2)]....\left(i\right)$$
Now, let us draw the table for trigonometric ratios of general angles.

From the above table, we can see that,
$$\mathrm{cosec}\left({\displaystyle \frac{\pi }{6}}\right)=2$$
$$\Rightarrow {\mathrm{cosec}}^{-1}\left(2\right)={\displaystyle \frac{\pi }{6}}$$
So, by substituting the value of $${\mathrm{cosec}}^{-1}\left(2\right)$$ in the expression (i), we get,
$$E={\sin }^{-1}[\cos (-2.{\displaystyle \frac{\pi }{6}})]$$
$$E={\sin }^{-1}[\cos (-{\displaystyle \frac{\pi }{3}})]$$
We know that, $$\cos (-\theta )=\cos \theta$$. By using this in the above expression, we get,
$$E={\sin }^{-1}[\cos {\displaystyle \frac{\pi }{3}}]$$
Now, from the trigonometric table, we can see that, $$\cos {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{1}{2}}$$. So, by substituting the value of $$\cos {\displaystyle \frac{\pi }{3}}$$ in the above expression, we get,
$$E={\sin }^{-1}\left({\displaystyle \frac{1}{2}}\right)....\left(ii\right)$$
Now we know that the range of principal value of $${\sin }^{-1}\left(x\right)$$ lies between $$[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$$.
From the table of general trigonometric ratios, we get,
$$\sin \left({\displaystyle \frac{\pi }{6}}\right)={\displaystyle \frac{1}{2}}$$
$$\Rightarrow {\sin }^{-1}\left({\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{\pi }{6}}$$
Now by substituting the value of $${\sin }^{-1}\left({\displaystyle \frac{1}{2}}\right)$$ in the expression (ii), we get,
$$E={\displaystyle \frac{\pi }{6}}$$
Hence, we get the value of $${\sin }^{-1}[\cos \left(2{\mathrm{cosec}}^{-1}(-2)\right)]$$ as $${\displaystyle \frac{\pi }{6}}$$.

Note: In this question, students must take care that the value of the angle must lie in the range of $${\mathrm{cosec}}^{-1}x$$ which is $$[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]-\left\{0\right\}$$ and $${\sin }^{-1}x$$ which is $$[{\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}]$$ accordingly. Also, students can verify their answer by equating the given expression with $${\displaystyle \frac{\pi }{6}}$$ and taking sin on both sides and keep solving until LHS = RHS.