# Proving Trig Identities

• Nov 12th 2008, 05:09 PM
peekaboos
Proving Trig Identities
I'm having troubles proving some trig identities. Any help is appreciated. :)

Question 1:
1 - sin2x / cos2x = 1 - tanx / 1 + tanx

Ignore question 2, I figured it out.

Question 2:
1 + sin2x / cosx + sinx = cos2x / cosx - sinx
• Nov 12th 2008, 07:45 PM
Soroban
Hello, peekaboos!

Quote:

1) Prove: .$\displaystyle \frac{1-\sin2x}{\cos2x} \;=\;\frac{1-\tan x}{1 + \tan x}$

The right side is: .$\displaystyle \frac{1-\tan x}{1 + \tan x} \;=\;\frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad \frac{{\color{blue}\cos x}\left(1 - \frac{\sin x}{\cos x}\right)} {{\color{blue}\cos x}\left(1 + \frac{\sin x}{\cos x}\right)} \;=\;\frac{\cos x - \sin x}{\cos x + \sin x}$

Multiply by $\displaystyle \frac{\cos x - \sin x}{\cos x - \sin x}\!: \quad \frac{\cos x - \sin x}{\cos x + \sin x}\cdot{\color{blue}\frac{\cos x - \sin x}{\cos x - \sin x}} \;=\;\frac{(\cos x - \sin x)^2}{\cos^2\!x - \sin^2\!x}$

. . $\displaystyle = \;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}} - \overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{\underbrace{\cos^2\!x - \sin^2\!x}_{\text{This is }\cos2x}} \;=\;\frac{1-\sin2x}{\cos2x}$