The value of $4 sin A {cos}^{3} A - 4 cos A {sin}^{3} A$ is equal to.

cos 2 A

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sin 3 A

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sin 2 A

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sin 4 A

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Solution

The correct option is D $sin 4 A$
$4 sin A {c o s}^{3} A - 4 cos A {sin}^{3} A$
$=$ $4 sin A cos A$ $[{cos}^{2} A - {sin}^{2} A]$
$=$ $2 [2 sin A cos A] [cos 2 A]$

$=$ $2 sin 2 A cos 2 A$

$=$ $sin 4 A$

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Similar questions

cos A + {cos}^{2} A = 1,

{sin}^{2} A + {sin}^{4} A

\frac{c o s 3 A - c o s A}{s i n 3 A - s i n A} + \frac{c o s 2 A - c o s 4 A}{s i n 4 A - s i n 2 A} =

\frac{sin A}{cos 2 A cos 3 A}

Prove that:

sin 4A = 4 sin A $c o s^{3}$ A - 4 cos A $s i n^{3}$ A

\frac{{cos}^{4} A}{{cos}^{2} B} + \frac{{sin}^{4} A}{{sin}^{2} B} = 1

\frac{{cos}^{4} B}{{cos}^{2} A} + \frac{{sin}^{4} B}{{sin}^{2} A}

{cos}^{4} A - {cos}^{2} A = {sin}^{4} A - {sin}^{2} A

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