Question

If $\cos \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha$, where $0<\alpha ,\beta <\pi /2$, then $\cos 2\beta$ is equal to

• A. $-2{\sin }^2(\frac{\pi }{4}-\alpha )$
• B. $-2{\cos }^2(\frac{\pi }{4}+\alpha )$
• C. $2{\sin }^2(\frac{\pi }{4}+\alpha )$
• D. $2{\cos }^2(\frac{\pi }{4}-\alpha )$

Answer 1

Answer: (A), (B)

The correct options are
A $-2{\sin }^2(\frac{\pi }{4}-\alpha )$
B $-2{\cos }^2(\frac{\pi }{4}+\alpha )$

Since $sin\alpha$, $cos\beta$, $cos\alpha$ are in geometric series so,
$co{s}^2\beta =sin\alpha \ast cos\beta$
$2\ast co{s}^2\beta =2\ast sin\alpha \ast cos\alpha =sin2\alpha$
$cos2\beta =sin2\alpha -1$ ...(i)
Option $A=-2si{n}^2(\frac{\pi }{4}-\alpha )=cos2(\frac{\pi }{4}-\alpha )-1=cos(\frac{\pi }{2}-2\alpha )-1=sin2\alpha -1$
From equation $\left(1\right)$ & above formula this option A correct.
option $B=-2co{s}^2(\frac{\pi }{4}+\alpha )=-cos2(\frac{\pi }{4}+\alpha )-1=-cos(\frac{\pi }{2}+2\alpha )-1=sin2\alpha -1$
Hence, options 'A' & 'B' are correct