Question

Let $f\left( x \right)={{e}^{{{\cos }^{-1}}sin\left( x+\dfrac{\pi }{3} \right)}}$ , then
A. $f\left( \dfrac{8\pi }{9} \right)={{e}^{\dfrac{5\pi }{18}}}$
B. $f\left( \dfrac{8\pi }{9} \right)={{e}^{\dfrac{13\pi }{18}}}$
C. $f\left( -\dfrac{7\pi }{4} \right)={{e}^{\dfrac{\pi }{12}}}$
D. $f\left( -\dfrac{7\pi }{4} \right)={{e}^{\dfrac{11\pi }{12}}}$

Answer 1

Problems of this type can be solved by assuming the entire term in the power of $e$ to be $\varphi$ . We will put $\dfrac{8\pi }{9}$ and $-\dfrac{7\pi }{4}$ in place $x$ and put the corresponding value of $\varphi$ in the main equation. Further simplifying the expressions, we will get the values of $f\left( \dfrac{8\pi }{9} \right)$ and $f\left( -\dfrac{7\pi }{4} \right)$ to get which one is the correct option among the given ones.

Complete step-by-step answer:
The given function we have is
$f\left( x \right)={{e}^{{{\cos }^{-1}}sin\left( x+\dfrac{\pi }{3} \right)}}$
For, further simplifying the above function we assume the term ${{\cos }^{-1}}sin\left( x+\dfrac{\pi }{3} \right)$ as indices of $e$ to be equal to $\varphi$
$\Rightarrow \varphi ={{\cos }^{-1}}sin\left( x+\dfrac{\pi }{3} \right).....\left( 1 \right)$
As, we know the formula of inverse trigonometry ${{\cos }^{-1}}\theta =\dfrac{\pi }{2}-{{\sin }^{-1}}\theta$ in the range of $\left[ 0,\pi \right]$ $$
So, we put $\dfrac{8\pi }{9}$ in place of $x$ in the equation $\left( 1 \right)$ and get
$\Rightarrow \varphi ={{\cos }^{-1}}sin\left( \dfrac{8\pi }{9}+\dfrac{\pi }{3} \right).....\left( 2 \right)$
Also, we can write the term $\dfrac{8\pi }{9}+\dfrac{\pi }{3}$ to be equal to $\dfrac{\pi }{2}+\theta$ as
$\Rightarrow \theta +\dfrac{\pi }{2}=\dfrac{8\pi }{9}+\dfrac{\pi }{3}$
Simplifying the above equation, we get the value of $\theta$ as
$\Rightarrow \theta =\dfrac{8\pi }{9}+\dfrac{\pi }{3}-\dfrac{\pi }{2}$
$\Rightarrow \theta =\dfrac{13\pi }{18}$
We rewrite the equation $\left( 2 \right)$ as
$\Rightarrow \varphi ={{\cos }^{-1}}sin\left( \dfrac{\pi }{2}+\dfrac{13\pi }{18} \right)$
As we know that $sin\left( \dfrac{\pi }{2}+\theta \right)=\cos \theta$ , rewriting the above equation as shown below
$\Rightarrow \varphi ={{\cos }^{-1}}\cos \left( \dfrac{13\pi }{18} \right)$
As we know ${{\cos }^{-1}}\cos \theta =\theta$ in the range of $\left[ 0,\pi \right]$ , rewriting the above equation we get
$\Rightarrow \varphi =\dfrac{13\pi }{18}$
Now, we put the value of $\varphi$ in the main function and get
$\Rightarrow f\left( \dfrac{8\pi }{9} \right)={{e}^{\dfrac{13\pi }{18}}}$
Again, putting $-\dfrac{7\pi }{4}$ in place of $x$ in equation $\left( 1 \right)$ we get
$\Rightarrow \varphi ={{\cos }^{-1}}sin\left( -\dfrac{7\pi }{4}+\dfrac{\pi }{3} \right)$
\Rightarrow \varphi ={{\cos }^{-1}}\left\\{ -sin\left( \dfrac{17\pi }{12} \right) \right\\}.....\left( 3 \right)
Also, we can rewrite the term $\dfrac{17\pi }{12}$ as $\dfrac{\pi }{2}+\dfrac{11\pi }{12}$ in equation $\left( 3 \right)$
\Rightarrow \varphi ={{\cos }^{-1}}\left\\{ -sin\left( \dfrac{\pi }{2}+\dfrac{11\pi }{12} \right) \right\\}
We rewrite the above equation as
\Rightarrow \varphi ={{\cos }^{-1}}\left\\{ -\cos \left( \dfrac{11\pi }{12} \right) \right\\}
Rewriting the above equation also as
\Rightarrow \varphi ={{\cos }^{-1}}\left\\{ -\cos \left( \pi -\dfrac{\pi }{12} \right) \right\\}
We know $\cos \left( \pi -\theta \right)=-\cos \theta$
Hence, \varphi ={{\cos }^{-1}}\left\\{ \cos \left( \dfrac{\pi }{12} \right) \right\\}
$\Rightarrow \varphi =\dfrac{\pi }{12}$
Now, we put the value of $\varphi$ in the main function and get
$\Rightarrow f\left( -\dfrac{7\pi }{4} \right)={{e}^{\dfrac{\pi }{12}}}$
Therefore, we conclude that both the options (B) and (C) are correct.

Note: While doing the conversions of inverse trigonometric functions we must be very careful about the range and the domains of the function as they vary with each other. Also, we must be careful about the $+$ and $-$ signs to avoid silly mistakes.