Question

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

Answer 1

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]}