trig proof

[sarashack]

Please help...

If a + b + c = pie, then prove that

tan(a) + tan(b) + tan(c) = tan(a) * tan(b) * tan(c)

I know this has something to do with the period of tangent being pie, but that's about as far as I've gotten.

Thanks!
Eagerly Awaiting Reply,
Sara

 

[jolly]

http://en.wikipedia.org/wiki/Pie

http://en.wikipedia.org/wiki/Pi

 

[sarashack]

pi vs. pie

Thanks for correcting me, jolly.

Do you have an answer to the trig problem?

 

[soroban]

Hello, Sara!

This is a classic (very old) problem . . .

If \(\displaystyle a\,+\,b\,+\,c\:=\:\pi\),
then prove that: \(\displaystyle \,\tan(a)\,+\,tan(b)\,+\,tan(c)\:=\:\tan(a)\cdot\tan(b)\cdot\tan(c)\)

We have:\(\displaystyle \,a\,+\,b\,+\,c\:=\:\pi\;\;\Rightarrow\;\;a\,+\,b\:=\:\pi\,-\,c\)


Take the tangent of both sides: \(\displaystyle \:\,\tan(a\,+\,b)\;=\;\tan(\pi\,-\,c)\)


Using compound-angle identities for tangent, we get:

\(\displaystyle \L\;\;\;\frac{\tan(a)\,+\,\tan(b)}{1\,-\,\tan(a)\cdot\tan(b)}\;=\;\frac{\tan(\pi)\,-\,\tan(c)}{1+\,\tan(\pi)\cdot\tan(c)}\)


Since \(\displaystyle \tan(\pi)\,=\,0\), we have: \(\displaystyle \L\,\frac{\tan(a)\,+\,\tan(b)}{1\,-\,\tan(a)\cdot\tan(b)}\)\(\displaystyle \;=\;-\tan(c)\)

And we have: \(\displaystyle \:\tan(a)\,+\,\tan(b)\;=\;-\tan(c)\,+\,\tan(a)\cdot\tan(b)\cdot\tan(c)\)

Therefore: \(\displaystyle \:\tan(a)\,+\,\tan(b)\,+\,\tan(c)\:=\:\tan(a)\cdot\tan(b)\cdot\tan(c)\)

 

[sarashack]

Thanks for helping me with this!!

I followed the first step, sort of like taking the log of both sides of an equation..

but I got lost on the second to the last step...'

Could you explain the rationale for adding {tan(a)*tan(b)*tan(c)} to the left side of the equation?

 

[skeeter]

Could you explain the rationale for adding {tan(a)*tan(b)*tan(c)} to the left side of the equation?

it wasn't "added" ... soroban multiplied both sides of the equation

[tan(a)+tan(b)]/[1 - tan(a)tan(b)] = -tan(c)

by the denominator of the left side ... [1 - tan(a)tan(b)]

{[tan(a)+tan(b)]/[1 - tan(a)tan(b)]}*[1 - tan(a)tan(b)] = -tan(c)*[1 - tan(a)tan(b)]

tan(a)+tan(b) = -tan(c) + tan(a)tan(b)tan(c)

add tan(c) to both sides and you're done.

 

[sarashack]

Thanks so very much to all of you for helping me with my problem!

"small minds discuss people, average minds discuss events, great minds discuss ideas"

--Sara