Deriving a Trig Identity

Date: 04/09/98 at 12:33:14
From: Jim
Subject: trig identity

We are trying to do the derivation of the identity:

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

for a triangle if x, y and z are angles of the triangle. We have tried 
to derive the formula and have searched in trig texts but have failed. 
This formula was used to solve a problem on a math contest sheet.

Date: 04/09/98 at 14:32:44
From: Doctor Jen
Subject: Re: trig identity

It's a lovely derivation - you'll kick yourself when you see! 

You need to make use of two results. First, if x, y and z are the 
angles in a triangle, clearly they add up to pi (working in radians). 
So z = pi-(x+y). 

Second, you need the identity:

                    tan A +- tan B
   tan (A+-B) =  -------------------
                 1 -+ (tan A)(tan B)

(+- is 'plus or minus', -+ is 'minus or plus').

Since z = pi-(x+y), tan z = tan (pi-(x+y)). So you can use the above 
identity, letting A be pi and B be (x+y). 

Work it through... you get tan z = -tan (x+y).

Use the identity again, to get:

              tan x + tan y
   tan z =  ------------------
            (tan x)(tan y) - 1

(I've skipped quite a lot of steps here, but you can work them out, 
I'm sure.)

Substitute this value for tan z into tan x + tan y + tan z,
then get everything over a common base, the base being 
(tan x)(tan y)-1 .

You should see that you then get:

                           (tan x)(tan y)(tan x + tan y)
   tan x + tan y + tan z = -----------------------------
                                (tan x)(tan y) - 1

But of course:

    (tan x + tan y)
   ------------------ = tan z, 
   (tan x)(tan y) - 1

and so you have the required result.

Think about where this formula might conceivably NOT work. Investigate 
if it's true for a right triangle, for example - you might want to 
check this out. 

-Doctor Jen,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   

Date: 04/20/98 at 06:22:54
From: James Yandell
Subject: Re: trig identity

Thank you.