Question

Let $f\left(x\right)=e^{{\cos }^{-1}\sin (x+\frac{\pi }{3})}$, then $f\left(\frac{8\pi }{9}\right)$ equals

• A. $e^{\frac{7\pi }{12}}$
• B. $e^{\frac{13\pi }{18}}$
• C. $e^{\frac{5\pi }{18}}$
• D. $e^{\frac{\pi }{12}}$

Answer 1

Answer: (B)

$e^{\frac{13\pi }{18}}$

$f\left(x\right)=e^{{\cos }^{-1}\sin (x+\frac{\pi }{3})}$
$\therefore f\left(\frac{8\pi }{9}\right)$

$f\left(x\right)=e^{{\cos }^{-1}\sin (\frac{\pi }{3}+\frac{8\pi }{9})}$
$=e^{{\cos }^{-1}\sin \left(\frac{11\pi }{9}\right)}$
$=e^{{\cos }^{-1}\sin \left(\frac{22\pi }{18}\right)}$
$=e^{{\cos }^{-1}\sin \left(\frac{9\pi }{18}+\frac{13\pi }{18}\right)}$
$=e^{{\cos }^{-1}\sin \left(\frac{\pi }{2}+\frac{13\pi }{18}\right)}$
$=e^{{\cos }^{-1}\cos \left(\frac{13\pi }{18}\right)}$
$=e^{\frac{13\pi }{18}}$