How do you simplify $1 - 4 {sin}^{2} x {cos}^{2} x$ $1-4sin^2xcos^2x$ ?

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How do you simplify $1 - 4 {sin}^{2} x {cos}^{2} x$ $1-4sin^2xcos^2x$ ?

1 Answer

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Answer 1 · 2015-09-24T18:55:04

It can be simplified to ${cos}^{2} (2 x)$ $cos^2(2x)$ or $\frac{1}{2} (cos 4 x + 1)$ $1/2(cos4x+1)$

Explanation:

We have that

$1 - 4 {sin}^{2} x \cdot {cos}^{2} x = (1^{2} - {(2 cos x \cdot sin x)}^{2}) = (1 - {(sin 2 x)}^{2}) = {cos}^{2} (2 x)$ $1-4sin^2x*cos^2x=(1^2-(2cosx*sinx)^2)=(1-(sin2x)^2)=cos^2(2x)$

Also we know that

$cos (4 x) = 2 \cdot {cos}^{2} (2 x) - 1 \Rightarrow {cos}^{2} (2 x) = \frac{1}{2} (cos 4 x + 1)$ $cos(4x)=2*cos^2(2x)-1=>cos^2(2x)=1/2(cos4x+1)$

Remarks

It is known that : ${sin}^{2} (2 x) + {cos}^{2} (2 x) = 1$ $sin^2(2x)+cos^2(2x)=1$

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How do you simplify $1 - 4 {sin}^{2} x {cos}^{2} x$ $1-4sin^2xcos^2x$ ?

1 Answer

Explanation:

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How do you simplify 1-4sin^2xcos^2x1−4sin2xcos2x1-4sin^2xcos^2x?

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How do you simplify $1 - 4 {sin}^{2} x {cos}^{2} x$ $1-4sin^2xcos^2x$ ?