Question

The value of expression $(1-\cos \frac{\pi }{10})(1+\cos \frac{3\pi }{10})(1+\cos \frac{7\pi }{10})(1-\cos \frac{9\pi }{10})$ is

• A. $\frac{1}{8}$
• B. $\frac{1}{16}$
• C. $\frac{1}{4}$
• D. 0

Answer 1

The correct option is B $\frac{1}{16}$

$(1-\cos \frac{\pi }{10})(1+\cos \frac{3\pi }{10})(1+\cos \frac{7\pi }{10})(1-\cos \frac{9\pi }{10})$

$=(1-\cos \frac{\pi }{10})(1+\cos \frac{3\pi }{10})(1-\cos \frac{3\pi }{10})(1+\cos \frac{\pi }{10})$

$=(1-{\cos }^2\frac{\pi }{10})(1-{\cos }^2\frac{3\pi }{10})$

$={\sin }^2\frac{\pi }{10}{\sin }^2\frac{3\pi }{10}$

$={\left(\sin \frac{\pi }{10}\sin \frac{3\pi }{10}\right)}^2$

$={\left(\sin {18}^o\sin {54}^o\right)}^2$

$={\left(\sin {18}^o\cos {36}^o\right)}^2$

$={(\frac{\sqrt{5}-1}{4}\cdot \frac{\sqrt{5}+1}{4})}^2$

$={\left(\frac{5-1}{16}\right)}^2$

$={\left(\frac{1}{4}\right)}^2$

$=\frac{1}{16}$.