How do you prove that $2 {cos}^{4} x + (2 - 2 {cos}^{2} x) {cos}^{2} x = cos 2 x + 1$ $2cos^4x + (2-2cos^2x)cos^2x = cos2x + 1$ ?

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How do you prove that $2 {cos}^{4} x + (2 - 2 {cos}^{2} x) {cos}^{2} x = cos 2 x + 1$ $2cos^4x + (2-2cos^2x)cos^2x = cos2x + 1$ ?

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Answer 1 · 2016-02-28T10:47:52

If you simplify the left hand side of the equation it is simply $2 {cos}^{2} x$ $2cos^2x$ .

It then becomes the double angle identity for $cos x$ $cosx$ .

$2 {cos}^{2} x = cos 2 x + 1$ $2cos^2x = cos2x + 1$

Explanation:

The double angle identity is a special case of the compound angle formula.

$cos (a + b) = cos (a) cos (b) - sin (a) sin (b)$ $cos(a+b) = cos(a)cos(b) - sin(a)sin(b)$

So letting $x = a = b$ $x = a = b$ , we get

$cos (2 x) = cos (x + x)$ $cos(2x) = cos(x+x)$

$= cos (x) cos (x) - sin (x) sin (x)$ $= cos(x)cos(x) - sin(x)sin(x)$

$= {cos}^{2} (x) - {sin}^{2} (x)$ $= cos^2(x) - sin^2(x)$

$= {cos}^{2} (x) - [1 - {cos}^{2} (x)]$ $= cos^2(x) - [1 - cos^2(x)]$

$= 2 {cos}^{2} (x) - 1$ $= 2cos^2(x) - 1$

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How do you prove that $2 {cos}^{4} x + (2 - 2 {cos}^{2} x) {cos}^{2} x = cos 2 x + 1$ $2cos^4x + (2-2cos^2x)cos^2x = cos2x + 1$ ?

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How do you prove that 2cos^4x + (2-2cos^2x)cos^2x = cos2x + 1 2cos4x+(2−2cos2x)cos2x=cos2x+1 2cos^4x + (2-2cos^2x)cos^2x = cos2x + 1 ?

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Explanation:

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How do you prove that $2 {cos}^{4} x + (2 - 2 {cos}^{2} x) {cos}^{2} x = cos 2 x + 1$ $2cos^4x + (2-2cos^2x)cos^2x = cos2x + 1$ ?