Answer:
\(2\sin^2 x\cos^2 x=\frac{1-(\cos 2x)^2}{2}\)
Step-by-step explanation:
Given : Expression - \(2\sin^2 x\cos^2 x\)
To simplify : The given expression?
Solution :
Expression - \(2\sin^2 x\cos^2 x\)
Apply the squared identity of the trigonometric function,
\(\sin^2 \theta=\frac{1-\cos 2\theta}{2}\)
\(\cos^2 \theta=\frac{1+\cos 2\theta}{2}\)
Substitute the value in the given expression,
\(=2\sin^2 x\cos^2 x\)
\(=2\times \frac{1-\cos 2x}{2}\times \frac{1+\cos 2x}{2}\)
\(=\frac{(1-\cos 2x)(1+\cos 2x)}{2}\)
Apply, \(a^2-b^2=(a+b)(a-b)\)
\(=\frac{1^2-(\cos 2x)^2}{2}\)
\(=\frac{1-(\cos 2x)^2}{2}\)
Therefore, \(2\sin^2 x\cos^2 x=\frac{1-(\cos 2x)^2}{2}\)
