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

1 Answer
Apr 14, 2016

Answer:

Use
#tan x=sinx/cos x, sin 2x = 2 sin x cos x and cos 2x = cos^2x-sin^2x#, 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 x/(cos^2x-sin^2x)#
#=(sin 2x)/(cos 2x)=tan 2x#

Proofs for #sin 2x = 2 sin x cos x and cos 2x = 1 - 2 sin^2x#:

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]}