If LHS is $= 4 ({cos}^{3} (10 °) + {sin}^{3} (20 °))$ , then RHS will be.

4 (cos (10 °) + sin (20 °))

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3 (cos (20 °) + sin (10 °))

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3 (cos (10 °) + sin (20 °))

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4 (cos (20 °) + sin (10 °))

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Solution

The correct option is C $3 (cos (10 °) + sin (20 °))$
$sin (3 θ) = 3 sin (θ) - 4 {sin}^{3} (θ)$

$cos (3 θ) = 4 {cos}^{3} (θ) - 3 cos (θ)$

$4 ({cos}^{3} (10 °) + {sin}^{3} (20 °)) = cos (30 °) + 3 cos (10 °) + 3 sin (20 °) - sin (60 °)$

$\Rightarrow \frac{\sqrt{3}}{2} + 3 (cos (10 °) + sin (20 °)) - \frac{\sqrt{3}}{2}$

$\Rightarrow 3 (cos (10 °) + sin (20 °))$

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Similar questions

= 4 ({cos}^{3} (10 °) + {sin}^{3} (20 °))

4 (\cos^{3} 10 ° + \sin^{3} 20 °) = 3 (\cos 10 ° + \sin 20 °)

Prove that following identities:

$4 (c o s^{3} 10^{\circ} + s i n^{3} 20^{\circ}) = 3 (c o s 10^{\circ} + s i n 20^{\circ})$

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