The Value of \( \frac{\sin 3x}{1+2\cos 2x} \) is Equal to What?
Question
Find the value of
\[ \frac{\sin 3x}{1+2\cos 2x} \]
(a) \(\cos x\)
(b) \(\sin x\)
(c) \(-\cos x\)
(d) \(\sin x\)
Solution
Use the identity
\[ \sin 3x=\sin x+2\sin x\cos 2x \]
\[ \sin 3x = \sin x(1+2\cos 2x) \]
Substituting into the given expression,
\[ \frac{\sin 3x}{1+2\cos 2x} = \frac{\sin x(1+2\cos 2x)} {1+2\cos 2x} \]
\[ = \sin x \]
(Provided \(1+2\cos 2x \neq 0\).)
Final Answer
\[ \boxed{\sin x} \]
Hence, the correct option is (b) \(\sin x\).