Question

If $\alpha ϵ[-2\pi ,2\pi ]$ and $\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}=\sqrt{2}(\cos {36}^{\circ }-\sin {18}^{\circ })$ then a value of $\alpha$ is

• A. $\frac{7\pi }{6}$
• B. $\frac{\pi }{6}$
• C. $-\frac{5\pi }{6}$
• D. $-\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{7\pi }{6}$

Using value of $sin{18}^{\circ }$ and $cos{36}^{\circ }$
$cos\frac{\alpha }{2}+sin\frac{\alpha }{2}=\sqrt{2}(\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}+1}{4})$
$\Rightarrow cos\frac{\alpha }{2}+sin\frac{\alpha }{2}=\sqrt{2}/2$
squaring both sides
$\Rightarrow 1+sin\alpha =1/2$ $\Rightarrow sin\alpha =-1/2$
Therefore general solution is, $\alpha =n\pi +(-1{)}^n\frac{7\pi }{6}$
in $[-2\pi ,2\pi ]$, the value of $\alpha$ is $7\pi /6$