Question

Let $f\left(\theta \right)={\sum }_{r=1}^9\left(sin\right(2r-1)\theta +cos2r\theta )andsin\frac{\pi }{18}=a,thenf\left(\frac{\pi }{18}\right)isequalto$

• A. $\frac{1+a}{a}$
• B. $\frac{a}{1+a}$
• C. $\frac{1-2a}{1+a}$
• D. $\frac{1-a}{a}$

Answer 1

Answer: (A)

$\frac{1+a}{a}$

Solve:

$\text{Given,}f\left(\theta \right)={\sum }_{r=1}^9\left(\sin \right(2r-1)\theta +\cos 2r\theta )$

$\sin \frac{\pi }{18}=a$

$f\left(\theta \right)=\sin \theta +\cos 2\theta +\sin 3\theta +\cos 4\theta +\sin 5\theta +$

$\cos 6\theta +\sin 17\theta +\cos 18\theta$

on seprating $\cos$ and $\sin$ term we get

$f\left(\theta \right)=(\sin \theta +\sin 3\theta +\sin 5\theta +\dots +\sin 17\theta )+(\cos 2\theta$

$+\cos 4\theta +\cos 6\theta +\dots \cos 18\theta )$

We know, that

$\sin \alpha +\sin (\alpha +\beta )+\sin (\alpha +2\beta )+\dots \sin (\alpha +(n-1\left)\beta \right)$

$=\frac{\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\sin (\alpha +\frac{(n-1)\beta }{2})$

and

$\cos \alpha +\cos (\alpha +B)+\cos (\alpha +2\beta )+\dots \cdot \cos (\alpha +(n-1\left)\beta \right)=\frac{\sin \frac{n\beta }{2}}{\sin \frac{B}{2}}\cos (\alpha +\frac{(n-1)\beta }{2})$

$\Rightarrow f\left(\theta \right)=\frac{\sin \frac{9\left(2\theta \right)}{2}}{\sin \left(\frac{20}{2}\right)}\sin \left(\frac{\theta +17\theta }{2}\right)+\frac{\sin 9\theta }{\sin \theta }\cos \left(10\theta \right)$

$\Rightarrow f\left(\theta \right)=\frac{\sin 9\theta }{\sin \theta }\sin 9\theta +\frac{\sin 9\theta }{\sin \theta }\cos \left(10\theta \right)$

$\Rightarrow f\left(\theta \right)=\frac{\sin 9\theta }{\sin \theta }[\sin 9\theta +\cos (10\theta \left)\right]$

$\Rightarrow f\left(\frac{\pi }{18}\right)=\frac{\sin \pi /2}{\sin \frac{\pi }{18}}[\sin \frac{\pi }{2}+\cos (\frac{\pi }{2}+\frac{\pi }{18})]$

$=\Rightarrow f\left(\frac{\pi }{18}\right)=\frac{1}{a}[1+\sin \frac{\pi }{18}]$

$\Rightarrow f\left(\frac{\pi }{18}\right)=\frac{1+a}{a}$