How do you prove that $2 {cos}^{4} x + 2 {sin}^{2} x {cos}^{2} x = cos 2 x + {sin}^{2} x + {cos}^{2} x$ $2cos^4x + 2sin^2xcos^2x = cos2x + sin^2x+cos^2x$ ?

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Answer 1 · 2018-07-15T04:15:24

Here ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$

The question provided by you now can be written as:-

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

$\Rightarrow$ $=>$ $2 {cos}^{4} x + 2 {sin}^{2} x {cos}^{2} x$ $2 cos^4x + 2sin^2xcos^2x$

$\Rightarrow$ $=>$ Now taking $2 {cos}^{2} x$ $2cos^2 x$ common we get,

$\Rightarrow$ $=>$ $2 {cos}^{x} ({cos}^{2} x + {sin}^{2} x)$ $2cos^x (cos^2x+sin^2x)$

$\Rightarrow$ $=>$ Now this equation becomes, $2 {cos}^{2} x$ $2cos^2x$

$\Rightarrow$ $=>$ This can be written as:- $cos 2 x + 1$ $cos 2x +1$

How do you prove that 2cos^4x + 2sin^2xcos^2x = cos2x + sin^2x+cos^2x 2cos4x+2sin2xcos2x=cos2x+sin2x+cos2x 2cos^4x + 2sin^2xcos^2x = cos2x + sin^2x+cos^2x ?