This is a question from AJC/C2/MY/P1/10. Quite challenging. But once you get the hang of it, you should be capable of solving other variations.

Using the formula for $sin(A+B)$, prove the $sin(3A)=3sinA-4sin^{3}A$.

(i) Hence show that $\sum_{r=1}^{n} \frac{4}{3^r} sin^{3}(3^{r}\theta)=sin(3\theta)-\frac{1}{3^n} sin(3^{n+1}\theta)$

(ii) Deduce the sum of the series $\sum_{r=1}^{n} sin^{2}(3^{r}\theta)cos(3^{r}\theta)$

Ans: $\frac{1}{4} [cos(3\theta)-cos(3^{n+1}\theta)]$