We have #160; sin 3x =3sin x 4sin ^ 3 x ⇒ 4sin ^ 3 x =3sin x sin 3x ⇒sin ^ 3 x = 3sin x sin 3x #160; 4 Use above formula #160; ...

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Principal Solution of Trigonometric Equation

Expand sin 3...

Question

Expand
${sin}^{3} (2 x + 1)$

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Solution

We have
$sin 3 x = 3 sin x - 4 {sin}^{3} x \Rightarrow 4 {sin}^{3} x = 3 sin x - sin 3 x \Rightarrow {sin}^{3} x = \frac{3 sin x - sin 3 x}{4}$

Use above formula
${sin}^{3} (2 x + 1)$

$= \frac{3 sin (2 x + 1) - sin 3 (2 x + 1)}{4}$

$= \frac{3 sin (2 x + 1) - sin (6 x + 3)}{4}$

$= \frac{3 (sin 2 x cos 1 + cos 2 x sin 1) - (sin 6 x cos 1 + cos 6 x sin 1)}{4}$

$= \frac{cos 1}{4} (3 sin 2 x - sin 6 x) + \frac{sin 1}{4} (3 cos 2 x - cos 6 x)$

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