How do you prove $cos 2 x = \frac{1 - {tan}^{2} (x)}{1 + {tan}^{2} (x)}$ $cos2x= (1-tan^2(x))/ (1+tan^2(x))$ ?

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How do you prove $cos 2 x = \frac{1 - {tan}^{2} (x)}{1 + {tan}^{2} (x)}$ $cos2x= (1-tan^2(x))/ (1+tan^2(x))$ ?

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Answer 1 · 2015-04-16T17:03:51

In this way:

(remembering that $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ and ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ ),

the second member becomes:

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

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

$= {cos}^{2} x - {sin}^{2} x$ $=cos^2x-sin^2x$ ,

that is the development of the formula of $cos 2 x$ $cos2x$ .

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How do you prove $cos 2 x = \frac{1 - {tan}^{2} (x)}{1 + {tan}^{2} (x)}$ $cos2x= (1-tan^2(x))/ (1+tan^2(x))$ ?

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