Question

If $\cos B$ is the geometric mean of $\sin A$ and $\cos A$, where $0<A,B<\frac{\pi }{2}$, then the value(s) of $\cos 2B$ is/are

• A. $1-\sin 2A$
• B. $2{\sin }^2(\frac{\pi }{4}-A)$
• C. $\sin 2A-1$
• D. $-2{\sin }^2(\frac{\pi }{4}-A)$

Answer 1

Answer: (C), (D)

$-2{\sin }^2(\frac{\pi }{4}-A)$

Given:
$(\cos B{)}^2=\sin A\cos A\Rightarrow 2(\cos B{)}^2=2\sin A\cos A\Rightarrow 2{\cos }^{2}B=\sin 2A\Rightarrow 1+\cos 2B=\sin 2A\therefore \cos 2B=\sin 2A-1\Rightarrow \cos 2B=-(1-\sin 2A)\Rightarrow \cos 2B=-(\cos A-\sin A{)}^2\Rightarrow \cos 2B=-2{(\frac{1}{\sqrt{2}}\cos A-\frac{1}{\sqrt{2}}\sin A)}^2\therefore \cos 2B=-2{\sin }^2(\frac{\pi }{4}-A)$