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Question
\[\frac{\sin 3x}{1 + 2 \cos 2x}\] is equal to
Options
cos x
sin x
– cos x
sin x
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Solution
sin x
\[\text{ We have } , \]
\[\frac{\sin 3x}{1 + 2\cos 2x} = \frac{3\text{ sin } x - 4 \sin^3 x}{1 + 2\left( 1 - 2 \sin^2 x \right)}\]
\[ = \frac{3\text{ sin } x - 4 \sin^3 x}{1 + 2 - 4 \sin^2 x}\]
\[ = \frac{\text{ sin } x\left( 3 - 4 \sin^2 x \right)}{\left( 3 - 4 \sin^2 x \right)}\]
\[ = \sin x\]
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