Question

Let $f\left(x\right)=e^{{\cos }^{-1}\sin (x+\pi /3)}$ Then

• A. $f\left(\frac{5\pi }{9}\right)=e^{5\pi /18}$
• B. $f\left(\frac{8\pi }{9}\right)=e^{13\pi /18}$
• C. $f(-\frac{7\pi }{4})=e^{\pi /12}$
• D. $f(-\frac{7\pi }{4})=e^{11\pi /12}$

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(\frac{8\pi }{9}\right)=e^{13\pi /18}$
C $f(-\frac{7\pi }{4})=e^{\pi /12}$

$f\left(x\right)=e^{{\cos }^{-1}\sin (x+\pi /3)}$
$f\left(x\right)=e^{{\cos }^{-1}\cos \left(\frac{\pi }{2}-(x+\frac{\pi }{3})\right)}$
$=e^{{\cos }^{-1}\cos \left(\frac{\pi }{6}-x\right)}$
$f\left(x\right)=e^{\left(\frac{\pi }{6}-x\right)}$
Now, we will check using options
Option A, $f\left(\frac{5\pi }{9}\right)=e^{\left(\frac{\pi }{6}-\frac{5\pi }{9}\right)}=e^{-\frac{7\pi }{18}}$
Option B, $f\left(\frac{8\pi }{9}\right)=e^{\left(\frac{\pi }{6}-\frac{8\pi }{9}\right)}=e^{-\frac{13\pi }{18}}$
Option C, $f\left(\frac{-7\pi }{4}\right)=e^{\left(\frac{\pi }{6}+\frac{7\pi }{4}\right)}=e^{\frac{23\pi }{12}}$