How do you prove ${sin}^{2} (x) {cos}^{2} (x) = \frac{1 - cos (4 x)}{8}$ $sin^2(x)cos^2(x) = (1-cos(4x))/8$ ?

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Answer 1 · 2015-05-14T23:05:07

Let's start form the Left Hand Side( $LHS$ $"LHS"$ ):

$\Rightarrow LHS = {sin}^{2} x {cos}^{2} x = {(\frac{2}{2} sin x cos x)}^{2}$ $=>" LHS "=sin^2xcos^2x =(2/2sinxcosx)^2$

Remember that, $2 sin x cos x = sin 2 x$ $2sinxcosx=sin2x$

$\Rightarrow LHS = {(\frac{1}{2} sin 2 x)}^{2} = \frac{1}{4} {sin}^{2} (2 x)$ $=>" LHS "=(1/2sin2x)^2 = 1/4sin^2(2x)$

Use the half angle identity, $cos 2 θ = 1 - 2 {sin}^{2} θ$ $cos2theta=1-2sin^2theta$

Replace $θ$ $theta$ by $2 x$ $2x$ $\Rightarrow cos 4 x = 1 - 2 {sin}^{2} (2 x)$ $=>cos4x=1-2sin^2(2x)$

Rearrange that $; {sin}^{2} (2 x) = \frac{1}{2} (1 - cos 4 x)$ $" ; "sin^2(2x)=1/2(1-cos4x)$

$\Rightarrow LHS = \frac{1}{4} (\frac{1}{2} (1 - cos 4 x))$ $=> " LHS"=1/4(1/2(1-cos4x))$

$= \frac{1}{8} (1 - cos 4) x = \frac{1 - cos (4 x)}{8} = RHS$ $=1/8(1-cos4)x=(1-cos(4x))/8=" RHS"$

And that's it!

How do you prove sin2(x)cos2(x)=1−cos(4x)8sin^2(x)cos^2(x) = (1-cos(4x))/8?