How do you prove the identity $\frac{1 - cos 2 x}{tan x} = sin 2 x$ $(1-cos2x) / tanx = sin2x$ ?

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How do you prove the identity $\frac{1 - cos 2 x}{tan x} = sin 2 x$ $(1-cos2x) / tanx = sin2x$ ?

1 Answer

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Answer 1 · 2015-10-01T01:21:13

See the explanation below.

Explanation:

Use one of the identities:

$cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x = cos^2x-sin^2x$
$cos 2 x = 1 - 2 {sin}^{2} x$ $cos2x = 1-2sin^2x$
$cos 2 x = 2 {cos}^{2} x - 1$ $cos2x = 2cos^2x-1$

Playing around with them on scratch paper (or thinking about them) will lead to using the second version.

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

$= \frac{2 {sin}^{2} x}{\frac{sin x}{cos x}}$ $= (2sin^2x)/(sinx/cosx)$

$= 2 {sin}^{2} x \cdot \frac{cos x}{sin x}$ $= 2sin^2x * cosx/sinx$

$= 2 sin x c o n s x$ $= 2sinxconsx$

$= sin 2 x$ $= sin2x$

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How do you prove the identity $\frac{1 - cos 2 x}{tan x} = sin 2 x$ $(1-cos2x) / tanx = sin2x$ ?

1 Answer

Explanation:

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How do you prove the identity $\frac{1 - cos 2 x}{tan x} = sin 2 x$ $(1-cos2x) / tanx = sin2x$ ?