Question

If $\sin \left(\log i\right)=a+ib,$ then $\cos \left(\log i^i\right)$ equals?

• A. -1
• B. 1
• C. 0
• D. Not defined

Answer 1

The correct option is C 0

$\sin \left(\log i^i\right)=a+ib$
$\Rightarrow a+ib=\sin \left(i\log i\right)=\sin \left(i\{\log \left|i\right|+iampi\}\right)$
$\sin \left(i\{\log 1+i\frac{\pi }{2}\}\right)=\sin (-\frac{\pi }{2})=-1+0i$
$\Rightarrow a=-1$ & $b=0$
$\therefore \sin \left(\log i^i\right)=-1$
$\Rightarrow \cos \left(\log i^i\right)=\sqrt{1-{\sin }^2\left(\log i^i\right)}=\sqrt{1-1}=0$