How do you prove $\frac{cos x - sin x}{cos x + sin x} = \frac{cos 2 x}{1 + sin 2 x}$ $(cosx-sinx)/(cosx+sinx)=(cos2x)/(1+sin2x)$ ?

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How do you prove $\frac{cos x - sin x}{cos x + sin x} = \frac{cos 2 x}{1 + sin 2 x}$ $(cosx-sinx)/(cosx+sinx)=(cos2x)/(1+sin2x)$ ?

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Answer 1 · 2015-11-26T22:41:06

Prove trig equation

Explanation:

$\frac{cos x - sin x}{cos x + sin x} =$ $(cos x - sin x)/(cos x + sin x) =$
$= [\frac{cos x - sin x}{cos x + sin x}] \frac{cos x + sin x}{cos x + sin x}] =$ $= [(cos x - sin x)/(cos x + sin x)] [cos x + sin x)/(cos x + sin x)] =$
$= \frac{{cos}^{2} x - {sin}^{2} x}{{(cos x + sin x)}^{2}} =$ $=(cos^2 x - sin^2 x)/(cos x + sin x)^2 =$
Apply trig identities:
${cos}^{2} x - {sin}^{2} x = cos 2 x$ $cos^2 x - sin^2 x = cos 2x$
${(cos x + sin x)}^{2} = {cos}^{2} x + 2 sin x . cos x + {sin}^{2} x = 1 + sin 2 x$ $(cos x + sin x)^2 = cos^2 x + 2sin x.cos x + sin^2 x = 1 + sin 2x$
Finally,
$\frac{cos x - sin x}{cos x + sin x} = \frac{cos 2 x}{1 + sin 2 x}$ $(cos x - sin x)/(cos x + sin x) = (cos 2x)/(1 + sin 2x)$

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How do you prove $\frac{cos x - sin x}{cos x + sin x} = \frac{cos 2 x}{1 + sin 2 x}$ $(cosx-sinx)/(cosx+sinx)=(cos2x)/(1+sin2x)$ ?

1 Answer

Explanation:

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How do you prove $\frac{cos x - sin x}{cos x + sin x} = \frac{cos 2 x}{1 + sin 2 x}$ $(cosx-sinx)/(cosx+sinx)=(cos2x)/(1+sin2x)$ ?