The Value of \( \cos^4x+\sin^4x-6\cos^2x\sin^2x \)
Question
Find the value of
\[ \cos^4x+\sin^4x-6\cos^2x\sin^2x \]
(a) \(\cos2x\)
(b) \(\sin2x\)
(c) \(\cos4x\)
(d) none of these
Solution
Use the identity
\[ \cos^4x+\sin^4x = (\cos^2x+\sin^2x)^2 – 2\cos^2x\sin^2x \]
\[ =1-2\cos^2x\sin^2x \]
Therefore,
\[ \cos^4x+\sin^4x-6\cos^2x\sin^2x = 1-8\cos^2x\sin^2x \]
Using
\[ \sin^22x = 4\sin^2x\cos^2x \]
\[ = 1-2\sin^22x \]
Now apply the identity
\[ \cos4x = 1-2\sin^22x \]
Hence,
\[ \cos^4x+\sin^4x-6\cos^2x\sin^2x = \cos4x \]
Final Answer
\[ \boxed{\cos4x} \]
Hence, the correct option is (c) \(\cos4x\).