Trig: Double-Angle Formulas

[kekepania]

I am attempting to study for my final and there are a few things that I was getting stuck on that I wanted to see if I could get help one.
I had one problem thant was:
prove: cos 2X = (1- tan^2 X/ 1+tan^2 X)
I then substitued with the identities and got

(1- (sin^2X/cos^2X))/(1+ (sin^2X/cos^2X))

I then replaced them with another identiy and got

(1- ((1-cos^2X)/(1+ sin^2X)))/(1+ ((1-cos^2X)/(1-sin^2X))

Then I got stuck. If anyone could help let me know.... thanks[/img]

 

[soroban]

Hello, kekepania!

You were almost there . . .

Prove: \(\displaystyle \:\cos 2X\:=\:\frac{1\,-\,\tan^2X}{1\,+\,\tan^2X}\)

I then substituted with the identities and got: \(\displaystyle \:\L\frac{1\,-\,\frac{\sin^2X}{\cos^2X}}{1\,+\,\frac{\sin^2X}{\cos^2X}}\;\;\) . . . good!

Multiply top and bottom by \(\displaystyle \cos^2X:\;\;\L\frac{\cos^2X\,-\,\sin^2X}{\cos^2X\,+\,\sin^2X}\;\;\) . . . got it?

The numerator is \(\displaystyle \cos2X\); the denominator is \(\displaystyle 1\).

 

[kekepania]

thank you for helping me!