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

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

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Answer 1 · 2018-04-30T03:13:08

$cos 2 x$ $cos2x$

Explanation:

Knowing that:
$cos 2 x = {cos}^{2} x - {sin}^{2} x = 2 {cos}^{2} x - 1$ $cos2x= cos^2x-sin^2x= 2cos^2x-1$
and ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

Start:
$\frac{{(2 {cos}^{2} x - 1)}^{2}}{{cos}^{4} x - {sin}^{4} x} =$ $(2cos^2x-1)^2/(cos^4x-sin^4x)=$

$\frac{{(cos 2 x)}^{2}}{({cos}^{2} x + {sin}^{2} x) ({cos}^{2} x - {sin}^{2} x)} =$ $(cos2x)^2/((cos^2x+sin^2x)(cos^2x-sin^2x))=$

$\frac{{(cos 2 x)}^{2}}{(1) ({cos}^{2} x - {sin}^{2} x)} =$ $(cos2x)^2/((1)(cos^2x-sin^2x))=$

$\frac{{(cos 2 x)}^{2}}{{cos}^{2} x - {sin}^{2} x} =$ $(cos2x)^2/(cos^2x-sin^2x)=$

$\frac{{(cos 2 x)}^{2}}{cos 2 x} =$ $(cos2x)^2/(cos2x)=$

$cos 2 x$ $cos2x$

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

1 Answer

Explanation:

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