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

Question

27045 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-26T18:17:13

Please see a Proof in the 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 · 2018-04-26T20:00:04

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:

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

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

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