Proving tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y)tan(y-z)tan(z-x)

[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 canceled 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.

 

[courtrigrad]

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

 

[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.

 

[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)]

 

[HallsofIvy]

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.

 

[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?

 

[Office_Shredder]

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

 

[Hurkyl]

I highly recommend doing a change of variable.

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

Because you've issued the (nonexistant) command

\tany

instead of the command

\tan

followed by a y.

(In other words, you need to use spaces)