Question

If $\sin \beta$ is geometric mean between $\sin \alpha$ and $\cos \alpha$ , then ${\cos }^{2}2\beta$ equals

• A. $2{\sin }^2(\frac{\pi }{4}-\alpha )2{\cos }^2(\frac{\pi }{4}+\alpha )$
• B. $2{\sin }^2(\frac{\pi }{3}-\alpha )2{\cos }^2(\frac{\pi }{3}+\alpha )$
• C. ${\sin }^2(\frac{\pi }{4}-\alpha )$ ${\cos }^2(\frac{\pi }{4}+\alpha )$
• D. ${\sin }^2(\frac{\pi }{3}-\alpha )$ ${\cos }^2(\frac{\pi }{3}+\alpha )$

Answer 1

Answer: (A)

$2{\sin }^2(\frac{\pi }{4}-\alpha )2{\cos }^2(\frac{\pi }{4}+\alpha )$

If $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha$, then ${\sin }^2\beta =\sin \alpha .\cos \alpha$
$\Rightarrow \cos 2\beta =1-2{\sin }^2\beta =1-2\sin \alpha .\cos \alpha \Rightarrow {\cos }^{2}2\beta ={(1-\sin 2\alpha )}^2$
Now if we simplify option A using $2\sin A.\cos B=\sin (A+B)+\sin (A-B)$, we get
$2{\sin }^2(\frac{\pi }{4}-\alpha )2{\cos }^2(\frac{\pi }{4}+\alpha )={\left(2\sin (\frac{\pi }{4}-\alpha )\cos (\frac{\pi }{4}+\alpha )\right)}^2={(\sin \frac{\pi }{2}+\sin (-2\alpha ))}^2={(1-\sin 2\alpha )}^2$
Similarly, we can verify other options.
Hence option A is correct.