Question

${\sin }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\sin }^2\frac{4\pi }{18}+{\cos }^2\frac{\pi }{18}+{\cos }^2\frac{2\pi }{18}+{\cos }^2\frac{4\pi }{18}$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

${\sin }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\sin }^2\frac{4\pi }{18}+{\cos }^2\frac{\pi }{18}+{\cos }^2\frac{2\pi }{18}+{\cos }^2\frac{4\pi }{18}$

$\Rightarrow$ $({\sin }^2\frac{\pi }{18}+{\cos }^2\frac{\pi }{18})+$ $({\sin }^2\frac{2\pi }{18}+{\cos }^2\frac{2\pi }{18})+$ $({\sin }^2\frac{4\pi }{18}+{\cos }^2\frac{4\pi }{18})$

We know that, ${\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow$ $1+1+1$

$\Rightarrow$ $3$

$\therefore$ ${\sin }^2\frac{\pi }{18}+{\sin }^2\frac{2\pi }{18}+{\sin }^2\frac{4\pi }{18}+{\cos }^2\frac{\pi }{18}+{\cos }^2\frac{2\pi }{18}+{\cos }^2\frac{4\pi }{18}=3$