The Value of \(2\cos x-\cos3x-\cos5x-16\cos^3x\sin^2x\)
Question
Find the value of
\[ 2\cos x-\cos3x-\cos5x-16\cos^3x\sin^2x \]
(a) \(2\)
(b) \(1\)
(c) \(0\)
(d) \(-1\)
Solution
Use the identity:
\[ \cos3x+\cos5x = 2\cos4x\cos x \]
Therefore,
\[ 2\cos x-\cos3x-\cos5x = 2\cos x-2\cos4x\cos x \]
\[ = 2\cos x(1-\cos4x) \]
Using
\[ 1-\cos4x=2\sin^22x \]
\[ = 4\cos x\sin^22x \]
Since
\[ \sin2x=2\sin x\cos x \]
\[ = 4\cos x(2\sin x\cos x)^2 \]
\[ = 16\cos^3x\sin^2x \]
Hence the given expression becomes
\[ 16\cos^3x\sin^2x – 16\cos^3x\sin^2x = 0 \]
Final Answer
\[ \boxed{0} \]
Hence, the correct option is (c) 0.