# Crazy tangent proof

1. Oct 20, 2006

### relskid

here's the proof:

$$\tan(x-y) + \tan(y-z) + \tan(z-x) = \tan(x-y)\tan(y-z)\tan(z-x)$$

in the chapter, we learned the sin, cos, tan addition laws, so i'm assuming that we're basically limited to use those and the fundamental trig functions such as $$sin^2x + cos^2x = 1$$ and the like.

i decided to work on the right side of the equation. i first converted the tan into sin/cos:

$$[\frac{\sin(x-y)}{\cos(x-y)}][\frac{\sin(y-z)}{\cos(y-z)}][\frac{\sin(z-x)}{\cos(z-x)}]$$

i worked out the numerator (the denominator is a real pain), and as you can imagine, i got a bunch of different sines and cosines. anywho, after stuff cancelled out and the like, i was left with:

$$\frac{-\sinx\sin^2z\cosx\cos^2y - \sin^2y\sinz\cos^2x\cosy + \siny\sin^2z\cos^2x\cosy - \sin^2x\siny\cosy\cos^2z + \sin^2x\sinz\cos^2y\cosz + \sinx\sin^2y\cosx\cosy\cosz}{\cos(x-y)\cos(y-z)\cos(z-x)}$$

right now, i'm stuck. if anyone can help, i'd be very grateful. :)

ps if the latex is not working, then i will just switch to the good old fashion messy look.

Last edited: Oct 20, 2006
2. Oct 20, 2006

Use the identity $$\tan(x-y) = \frac{\tan x - \tan y}{1+\tan x \tan y}$$ and work from the left side.

Last edited: Oct 20, 2006
3. Oct 20, 2006

### neutrino

If you know the identities for $$\sin\left(x\pm y\right)$$ and $$\cos\left(x\pm y\right)$$, why not convert the LHS to sin and cos and see what it simplifies to.

Edit: Convert to sin and cos AFTER you use the identity provided by the courtigrad.

4. Oct 21, 2006

### relskid

ok...

edit: i'm not getting the coding right, so i'm just going to write it all out.

[(sin^2x/cos^2x)(sinz/cosz) - (siny/cosy) - (sin^2y/cos^2y)(sinz/cosz) - (sinx/cosx)(sin^2y/cos^2y)(sinz/cosz) + (sinx/cosx)(sin^2y/cos^2y) - (sinx/cosx)(sin^2z/cos^2z) + (siny/cosy)(sin^2z/cos^2z) + (sinx/cosx)(sin^2y/cos^2y)(sin^2z/cos^2z) - (sin^2x/cos^2x)(siny/cosy)]

all divided by

[1 + (siny/cosy)(sinz/cosz) + (sinx/cosx)(siny/cosy) + (sinx/cosx)(sin^2y/cos^2y)(sinz/cosz) + (sinx/cosx)(siny/cosy)(sin^2z/cos^2z) + (sin^2x/cos^2x)(siny/cosy)(sinz/cosz) + (sin^2x/cos^2x)(sin^2y/cos^2y)(sin^2z/cos^2z)]

Last edited: Oct 21, 2006
5. Oct 21, 2006

### HallsofIvy

Staff Emeritus
You said that you know the tan addition law:
$$tan(a+ b)= \frac{tan(a)+ tan(b)}{1- tan(a)tan(b)}$$
so use that- don't go back to sine and cosine!

Of course, replacing b by -b, and remembering that tan(-b)= -tan(b),
$$tan(a-b)= \frac{tan(a)- tan(b)}{1+ tan(a)tan(b)}$$
as Coutrigrad said. Use that three times.

6. Oct 21, 2006

### relskid

i was just doing what neutrino suggested. :P

i have it all still in tangent written on paper, but i'm still not seeing anything that can be used.

anywho... converted back to tangent. :(

$$\frac{(\tan^2 x \tan z) - (\tan y) - (\tan^2 y \tan z) - (\tan x \tan^2 y \tan z) + (\tan x \tan^2 y) - (\tan x \tan^2 z) + (\tan y \tan^2 z) + (\tan x \tan^2 y \tan^2 z) - (\tan^2 x \tan y)}{1 + (\tan y \tan z) + (\tan x \tan y) + (\tan x \tan^2 y \tan z) + (\tan x \tan y \tan^2 z) + (\tan^2 x \tan y \tan z) + (\tan^2 x \tan^2 y \tan^2 z)}$$

edit: can anyone explain to me why some of the terms are not showing up?

Last edited: Oct 21, 2006
7. Oct 21, 2006

### Office_Shredder

Staff Emeritus
Now do the same thing for the right hand side, and you can check to see how you want to re-arrange the terms you got out of the left hand side to show it equals the right hand side

8. Oct 21, 2006

### Hurkyl

Staff Emeritus
I highly recommend doing a change of variable.

Because you've issued the (nonexistant) command

\tany