### Thinking [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ #5

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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 $\sum_{r=1}^n \text{sin}^2 (3^r \theta) \text{cos} (3^r \theta)$