Prove: $sin 3 a = 3 sin a - 4 {sin}^{3} a$ , when $a = 30^{\circ}$

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Solution

Given: $sin 3 a = 3 sin a - 4 {sin}^{3} a$ , when $a = 30 °$
Now,
LHS:
$\Rightarrow sin 3 a = sin 3 \times 30^{\circ} = sin 90^{\circ} = 1$ [Value of $sin 90^{\circ} = 1$ ]

RHS:
$3 sin a - 4 {sin}^{3} a = 3 sin 30^{\circ} - 4 {sin}^{3} 30^{\circ}$ [ Value of $sin 30^{\circ} = \frac{1}{2}$ ]
$\Rightarrow 3 \times (\frac{1}{2}) - 4 \times (\frac{1}{2})^{3}$
$= \frac{3}{2} - \frac{4}{8}$
$= \frac{3}{2} - \frac{1}{2}$
$= \frac{(3 - 1)}{2}$
$= \frac{2}{2}$
$= 1$
Hence, LHS = RHS

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