Prove that following identities-sin 3 A -sin 32 -3- A -sin 34 -3- A

Sin^3 A + sin^3 (2 π/3+A )+ sin^3 (4 π/3 + A ) {we know that} sin^3 A =3 sin A sin 3A/4 = (3 sinA sin 3A/4 )+{3 sin (2 π/3+A ) sin 3 (2 π/3+A )/4}+ {3 si ...

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Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

Prove that fo...

Question

Prove that following identities:

$s i n^{3} A + s i n^{3} (\frac{2 π}{3} + A) + s i n^{3} (\frac{4 π}{3} + A)$

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Solution

$s i n^{3} A + s i n^{3} (\frac{2 π}{3} + A) + s i n^{3} (\frac{4 π}{3} + A)$ ${we know that} s i n^{3} A = \frac{3 s i n A - s i n 3 A}{4}$

$= (\frac{3 s i n A - s i n 3 A}{4}) + {\frac{3 s i n (\frac{2 π}{3} + A) - s i n 3 (\frac{2 π}{3} + A)}{4}} + {\frac{3 s i n (\frac{4 π}{3} + A) - s i n 3 (\frac{4 π}{3} + A)}{4}} = [\frac{3 s i n A - s i n 3 A}{4}] + {\frac{3 s i n [π (\frac{2 π}{3} + A)] - s i n (2 π + 3 A)}{4}} + {\frac{3 s i n [π + (\frac{π}{3} + A)] - s i n (4 π + 3 A)}{4}}$

$= \frac{1}{4} {[3 s i n A - s i n 3 A] + [3 s i n (\frac{π}{3} - A) - s i n 3 A] - [3 s i n (\frac{π}{3} + A) + s i n 3 A]} = \frac{1}{4} [3 s i n A - s i n 3 A + 3 s i n (\frac{π}{3} - A) - 3 s i n (\frac{π}{3} + A) - s i n 3 A - s i n 3 A] = \frac{1}{4} [3 s i n A - 3 s i n 3 A + 3 (s i n (\frac{π}{3} - A) - s i n (\frac{π}{3} + A))] = \frac{1}{4} [3 s i n A - 3 s i n 3 A + 3 {2 c o s \frac{\frac{π}{3} - A + \frac{π}{3} + A}{2} s i n \frac{\frac{π}{3} - A - \frac{π}{3} - A}{2}}]$

$= \frac{1}{4} [3 s i n A - 3 s i n 3 A + 6 c o s \frac{π}{3} s i n (- A)] = \frac{1}{4}$ [3 sin A - 3 sin 3A - 3 sin A]

$= - \frac{3}{4} s i n 3 A = R H S$

LHS = RHS

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