Question

The value of ${\cos }^3\frac{\pi }{8}\cos \frac{3\pi }{8}+{\sin }^3\frac{\pi }{8}\sin \frac{3\pi }{8}$ is :

• A. $\frac{1}{4}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{2\sqrt{2}}$

${\cos }^3\frac{\pi }{8}\cos \frac{3\pi }{8}+{\sin }^3\frac{\pi }{8}\sin \frac{3\pi }{8}$

$={\cos }^3\frac{\pi }{8}[4{\cos }^3\frac{\pi }{8}-3\cos \frac{\pi }{8}]+{\sin }^3\frac{\pi }{8}[3\sin \frac{\pi }{8}-4{\sin }^3\frac{\pi }{8}]$

$=4[{\cos }^6\frac{\pi }{8}-{\sin }^6\frac{\pi }{8}]+3[{\sin }^4\frac{\pi }{8}-{\cos }^4\frac{\pi }{8}]$

$=4[{\cos }^2\frac{\pi }{8}-{\sin }^2\frac{\pi }{8}][{\cos }^4\frac{\pi }{8}+{\sin }^4\frac{\pi }{8}+{\cos }^2\frac{\pi }{8}{\sin }^2\frac{\pi }{8}]-3[{\cos }^2\frac{\pi }{8}-{\sin }^2\frac{\pi }{8}]$

$=[{\cos }^2\frac{\pi }{8}-{\sin }^2\frac{\pi }{8}][4(1-{\cos }^2\frac{\pi }{8}{\sin }^2\frac{\pi }{8})-3]$
$=\cos \frac{\pi }{4}[1-{\sin }^2\frac{\pi }{4}]=\frac{1}{2\sqrt{2}}$