Question

If $\sin \alpha$, 1 and $\cos 2\alpha$ are in GP then $\alpha$ is equal to

• A. $n\pi +{(-1)}^n\frac{\pi }{2},nϵZ$
• B. $n\pi +{(-1)}^{n-1}\frac{\pi }{2},nϵZ$
• C. $2n\pi ,nϵZ$
• D. none of these

Answer 1

Answer: (B)

$n\pi +{(-1)}^{n-1}\frac{\pi }{2},nϵZ$

They are in GP, then
$\sin \alpha .\cos 2\alpha =1$
$\sin \alpha (1-2{\sin }^2\alpha )=1$
$\sin \alpha -2{\sin }^3\alpha -1=0$
$\sin \alpha =-1$ or $\frac{1-i}{2}$ or $\frac{1+i}{2}$
So,
$\alpha =n\pi +(-1{)}^n\frac{-\pi }{2}$
$\alpha =n\pi +(-1{)}^{n-1}\frac{\pi }{2}$