IIT-JEE 2009 · Paper 1, Mathematics question 27

Sum of sines. $\displaystyle\sum_{m=1}^{15}\sin(2m-1)\theta=\frac{\sin^2 15\theta}{\sin\theta}$ (multiply by $2\sin\theta$ and telescope: $2\sin\theta\sin(2m-1)\theta=\cos(2m-2)\theta-\cos2m\theta$, total $1-\cos30\theta=2\sin^215\theta$).

At $\theta=2°$: $\dfrac{\sin^230°}{\sin2°}=\dfrac1{4\sin2°}$.

(D)