The Value of \( \frac{\cos 3x}{2\cos 2x-1} \) is Equal to What?
Question
Find the value of
\[ \frac{\cos 3x}{2\cos 2x-1} \]
(a) \(\cos x\)
(b) \(\sin x\)
(c) \(\tan x\)
(d) none of these
Solution
Use the identity
\[ \cos 3x = 4\cos^3x – 3\cos x \]
and
\[ \cos 2x = 2\cos^2x – 1 \]
Therefore,
\[ 2\cos 2x – 1 = 2(2\cos^2x-1)-1 \]
\[ = 4\cos^2x-3 \]
Now,
\[ \cos 3x = 4\cos^3x-3\cos x \]
\[ = \cos x(4\cos^2x-3) \]
Hence,
\[ \frac{\cos 3x}{2\cos 2x-1} = \frac{\cos x(4\cos^2x-3)} {4\cos^2x-3} \]
\[ = \cos x \]
(Provided \(4\cos^2x-3 \neq 0\).)
Final Answer
\[ \boxed{\cos x} \]
Hence, the correct option is (a) \(\cos x\).