Question

If $270^\circ < \theta < 360^\circ $ , then $\sqrt{2 + \sqrt{2 + 2 cos \theta }}$ is equal to

• A. $-2sin\left(\frac{\theta }{4}\right)$
• B. $2sin\left(\frac{\theta }{4}\right)$
• C. $2cos\left(\frac{\theta }{4}\right)$
• D. $-2cos\left(\frac{\theta }{4}\right)$

Answer 1

Answer: (B)

$\sqrt{2 + \sqrt{2 + 2 cos \theta }}=\sqrt{2 + \sqrt{2 \cdot 2 cos^{2} \frac{\theta }{2}}}$
$=\sqrt{2 + 2 \left|cos \frac{\theta }{2}\right|},135^\circ < \frac{\theta }{2} < 180^\circ $
$=\sqrt{2 - 2 cos \frac{\theta }{2}}=\left|2 sin \frac{\theta }{4}\right|$ ,
$67\frac{1 ^\circ }{2} < \frac{\theta }{4} < 90^\circ =2sin\frac{\theta }{4}$