Question

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

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

Answer 1

The correct option is A $\frac{1}{8}$

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

$=(1+\cos \frac{\pi }{10})[1+\cos \left(\pi \frac{7\pi }{10}\right)](1+\cos \frac{7\pi }{10})[1+\cos (\pi -\frac{\pi }{10})]$

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

$=(1-{\cos }^2\frac{\pi }{10})(1-{\cos }^2\frac{7\pi }{10})$

$={\sin }^2\frac{\pi }{10}\times {\sin }^2\frac{7\pi }{10}={\sin }^2{18}^o\times {\sin }^2{126}^o=(0.31{)}^2\times (0.81{)}^2$

$=0.063$

$=\frac{1}{8}$