Question

The value of $\cos ({\cos }^{-1}(-\frac{\sqrt{3}}{2})+\frac{\pi }{4})$ is?

• A. $-\frac{\sqrt{3}+1}{2\sqrt{2}}$
• B. $+\frac{\sqrt{3}+1}{2\sqrt{2}}$
• C. $-\frac{\sqrt{3}-1}{2\sqrt{2}}$
• D. $+\frac{\sqrt{3}-1}{2\sqrt{2}}$

Answer 1

The correct option is A $-\frac{\sqrt{3}+1}{2\sqrt{2}}$

$\because {\cos }^{-1}\theta \theta \in [0,\pi ]$
$\therefore {\cos }^{-1}(-\frac{\sqrt{3}}{2})=\frac{5\pi }{6}$
$\therefore \cos \left({\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\frac{\pi }{4}\right)=\cos (\frac{5\pi }{6}+\frac{\pi }{4})$
$=\cos \frac{5\pi }{6}\cos \frac{\pi }{4}+\sin \frac{5\pi }{6}\sin \frac{\pi }{4}$
$=\frac{-\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}=\frac{-\sqrt{3}+1}{2\sqrt{2}}$