Question

Solve ${\cos }^2\frac{\pi }{12}{\cos }^2\frac{\pi }{4}{\cos }^2\frac{5\pi }{12}$

• A. $\frac{1}{32}$
• B. $\frac{3-\sqrt{3}}{2}$
• C. $\frac{2}{3}$
• D. $\frac{2}{3+\sqrt{3}}$

Answer 1

The correct option is A $\frac{1}{32}$

${\cos }^2\frac{\pi }{12}={\cos }^2(\frac{\pi }{4}-\frac{\pi }{6})$

$⟹(\cos \frac{\pi }{4}\cos \frac{\pi }{6}+\sin \frac{\pi }{4}\sin \frac{\pi }{6}{)}^2$

$⟹(\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}{)}^2=\frac{(\sqrt{3}+1{)}^2}{8}$

Similarily, Putting ${\cos }^2\frac{5\pi }{12}={\cos }^2(\frac{\pi }{4}+\frac{\pi }{6})$ , we get $\frac{(\sqrt{3}-1{)}^2}{8}$

Atp, we have

$\frac{(\sqrt{3}-1{)}^2}{8}\times {\cos }^2\frac{\pi }{4}\times \frac{(\sqrt{3}+1{)}^2}{8}$

$\frac{\left(\right(\sqrt{3}+1\left)\right(\sqrt{3}-1){)}^2}{\mathrm{8.8.2}}$

$⟹\frac{({\sqrt{3}}^2-1{)}^2}{\mathrm{8.8.2}}$

$⟹\frac{4}{\mathrm{8.8.2}}$

$⟹\frac{1}{32}$