Question

How to verify Cos2x/(1+sin2x)=tan(pi/4-x)?

Answer 1

Please see a Proof in the Explanation.

Explanation:

`(cos2x)/(1+sin2x)`,

`=(cos^2x-sin^2x)/{(cos^2x+sin^2x)+2sinxcosx}`,

`={(cosx+sinx)(cosx-sinx)}/(cosx+sinx)^2`,

`=(cosx-sinx)/(cosx+sinx)`,

`={cosx(1-sinx/cosx)}/{cosx(1+sinx/cosx)}`,

`=(1-tanx)/(1+tanx)`,

`={tan(pi/4)-tanx}/{1+tan(pi/4)*tanx} quad` [because `tan(pi/4)=1`],

`=tan(pi/4-x)`, as desired!

Answer 2

First we remind ourselves `cos(2x)=cos (x+x)=cos^2x - sin^2x` and `sin(2x) = 2 sin x cos x`. Now let's approach from the other side.

`tan(pi/4 -x ) = {tan(pi/4) - tan x } / {1 + tan(pi/4) tan x}`

`= {1 - sin x/cos x} / {1 + sin x/cos x}`

`= {cos x - sin x}/{cos x + sin x}`

We know `cos 2x=cos^2x - sin^2 x` so our move is:

`= {cos x - sin x}/{cos x + sin x} cdot {cos x + sin x}/{cos x + sin x}`

`= { cos^2 x - sin^2 x}/{cos^2x + 2 cos x sin x + sin^2 x }`

`= {cos(2x) }/{1 + sin(2x)} quad sqrt`