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# How do you prove tan(2x) = (2tan(x))/(1-tan^2 x)?

Apr 14, 2016

Use
$\tan x = \sin \frac{x}{\cos} x , \sin 2 x = 2 \sin x \cos x \mathmr{and} \cos 2 x = {\cos}^{2} x - {\sin}^{2} x$, for the right hand side expression

#### Explanation:

2 tan x/(1-tan^2x)=(2sin x/cos x)/(1-(sin^2x/cos^2x)

$= 2 \sin x \cos \frac{x}{{\cos}^{2} x - {\sin}^{2} x}$
$= \frac{\sin 2 x}{\cos 2 x} = \tan 2 x$

Proofs for $\sin 2 x = 2 \sin x \cos x \mathmr{and} \cos 2 x = 1 - 2 {\sin}^{2} x$:

Use

Area of a $\triangle$ABC = 1/2(base)(altitude) = 1/2 bc sin A. Here,

it is the $\triangle$ ABC of a unit circle, with center at A, B and C on

the circle and $\angle$A = 2x. Here, AB = AC = 1, BC = 2 sinx and

the altitude from A, AN = cos x.

Area of $\triangle$ABC = 1/2 (BC) (AN) = 1/2(OA)(OB) sin A.

Readily, 1/2(2 sin x) cos x = 1/2 (1)(1)sin 2x.

Use ${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$, to get the cosine formula.

graph{(x^2+y^2-1)(y-x)(y+x)(x-0.707)=0[0 3 -0.73 0.73]}