Question

If $\pi <2\theta <\frac{3\pi }{2}$, then $\sqrt{2+\sqrt{2+2cos4\theta }}$ equals

• A. $-2cos\theta$
• B. $-2sin\theta$
• C. $2cos\theta$
• D. $2sin\theta$

Answer 1

Answer: (D)

$2sin\theta$

$\sqrt{2+\sqrt{2+2cos4\theta }}$

$=\sqrt{2+\sqrt{4co{s}^{2}2\theta }}$ [$\because \cos 2x=2{\cos }^{2}x-1$]

$=\sqrt{2+2|cos2\theta |}$

Now

$\pi <2\theta <\frac{3\pi }{2}$

Hence

$\frac{\pi }{2}<\theta <\frac{3\pi }{4}$.

Hence it is in the second quadrant.

$|cos2\theta |<0$.

Hence

$\sqrt{2-2cos2\theta }$

$=\sqrt{4si{n}^2\theta }$ $(\because$ $1-cos2x=2si{n}^{2}x)$

$=2sin\theta$. since $\frac{\pi }{4}<\theta <\frac{3\pi }{8}$.