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Question

To Prove : -
${sin}^{3} x + s i n^{3} (\frac{2 π}{3} + x) + s i n^{3} (\frac{4 π}{3} + x)$ = $- \frac{3}{4} s i n 3 x$

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Solution

L.H.S.= ${sin}^{3} x + s i n^{3} (\frac{2 π}{3} + x) + s i n^{3} (\frac{4 π}{3} + x)$ = $\frac{1}{4} (3 s i n x - s i n 3 x) + \frac{1}{4} (3 s i n (\frac{2 π}{3} + x) - s i n 3 (\frac{2 π}{3} + x)) + \frac{1}{4} (3 s i n (\frac{4 π}{3} + x) - s i n 3 (\frac{4 π}{3} + x))$
[Since, ${sin}^{3} x$ = $\frac{1}{4} (3 s i n x - s i n 3 x)$ ]
= $\frac{1}{4} [3 s i n x - s i n 3 x + 3 s i n (\frac{2 π}{3} + x) - s i n (2 π + 3 x) + 3 s i n (\frac{4 π}{3} + x) - s i n (4 π + 3 x)]$
= $\frac{1}{4} [3 s i n x - s i n 3 x + 3 s i n (\frac{2 π}{3} + x) - s i n (3 x) + 3 s i n (\frac{4 π}{3} + x) - s i n (3 x)]$
= $\frac{1}{4} [3 s i n x - 3 s i n 3 x + 3 (2 s i n (\frac{2 π + 2 x}{2}) c o s (\frac{2 π}{6})]$
[ Since, $(s i n A + s i n B) = 2 s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2})$ ]
= $\frac{1}{4} [3 s i n x - 3 s i n 3 x + 3 (- 2 s i n x (\frac{1}{2})]$
= $\frac{1}{4} [3 s i n x - 3 s i n 3 x - 3 s i n x]$
= $- \frac{3}{4} s i n 3 x$
= R.H.S.

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