Question #f236b

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Question #f236b

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Answer 1 · 2016-12-19T21:04:16

$cos 2 x$ $cos2x$

Explanation:

Use the following identities to solve the problem

• $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sintheta/costheta$

• ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta + sin^2theta = 1$

• $cos 2 θ = {cos}^{2} θ - {sin}^{2} θ$ $cos2theta = cos^2theta - sin^2theta$

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

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

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

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

$= cos 2 x$ $= cos2x$

Hopefully this helps!

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Question #f236b

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