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This is a very challenging trigonometry question from AJC Mid Years P1 Q10.

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

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

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

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