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

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Answer 1 · 2018-04-26T18:17:13

Please see a Proof in the Explanation.

$\frac{cos 2 x}{1 + sin 2 x}$ $(cos2x)/(1+sin2x)$ ,

$= \frac{{cos}^{2} x - {sin}^{2} x}{({cos}^{2} x + {sin}^{2} x) + 2 sin x cos x}$ $=(cos^2x-sin^2x)/{(cos^2x+sin^2x)+2sinxcosx}$ ,

$= \frac{(cos x + sin x) (cos x - sin x)}{{(cos x + sin x)}^{2}}$ $={(cosx+sinx)(cosx-sinx)}/(cosx+sinx)^2$ ,

$= \frac{cos x - sin x}{cos x + sin x}$ $=(cosx-sinx)/(cosx+sinx)$ ,

$= \frac{cos x (1 - \frac{sin x}{cos x})}{cos x (1 + \frac{sin x}{cos x})}$ $={cosx(1-sinx/cosx)}/{cosx(1+sinx/cosx)}$ ,

$= \frac{1 - tan x}{1 + tan x}$ $=(1-tanx)/(1+tanx)$ ,

$={tan(pi/4)-tanx}/{1+tan(pi/4)*tanx} quad$ [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:

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

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

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