Deriving a Trig IdentityDate: 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. |
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