Proving a+b+c=abc with Inverse Trig Functions

[ebola_virus]

a teacher wrote on the board

tan-1(a) + tan-1(b) + tan-1(c) = pi
[they are inverse trig functions btw, not the recipricol 1/tanx = cotx]

hence prove that

a + b + c = abc

wow. do you have any idea where i can start? thanks. I've been uttered clueless.

 

[siddharth]

You should first find what
\tan^{-1} x + \tan^{-1} y is.That is, can you find the (??) in the equation below?
\tan^{-1} x + \tan^{-1} y = \tan^{-1} (??)
Finding the above is very interesting.
You have to consider the cases when

xy<1
xy>1 and x,y >0
xy>1 and x,y <0

 

[ebola_virus]

where did x and y come from agn?

 

[siddharth]

x and y are just two variables. You can replace them with a and b if you wish.

 

[ComputerGeek]

a teacher wrote on the board
tan-1(a) + tan-1(b) + tan-1(c) = pi
[they are inverse trig functions btw, not the recipricol 1/tanx = cotx]
hence prove that
a + b + c = abc
wow. do you have any idea where i can start? thanks. I've been uttered clueless.

so you mean:
<br /> arctan(a) + arctan(b) + arctan(c) = \pi<br />

 

[Tide]

If the first statement is true then a, b and c can be represented by the interior angles of a triangle. There is no triangle for which a + b + c = a \cdot b \cdot c because the maximum possible value of a \cdot b \cdot c is (\pi/3)^3 which is less than \pi.

 

[NateTG]

If the first statement is true then a, b and c can be represented by the interior angles of a triangle. There is no triangle for which a + b + c = a \cdot b \cdot c because the maximum possible value of a \cdot b \cdot c is (\pi/3)^3 which is less than \pi.

An example solution is:
a=b=c=\sqrt{3}
so
\tan^{-1}(a)=\tan^{-1}(b)=\tan^{-1}(c)=\frac{\pi}{3}[/itex]<br /> Then<br /> abc=3 \sqrt{3}<br /> and<br /> a+b+c= 3 \sqrt{3}

 

[Tide]

Yikes! What was I thinking!

Thanks for catching that, Nate!

 

[ebola_virus]

er, i guess i got it

hehe i got it that makes more sort of. systematic sense.

using the double angle for tan thing

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

state that tan (a) = A
tan (b) = B you'll see why later.

anyways take the arctan of both sides of the double identity for tan and you get

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

now becuase tan (a) = A
a = arctan A and vice vresa for B

hence you end up wiht the arctan identity

arctan (A) + arctan (B) = arctan [(A+B)/(1-AB)]

and then you use that for a, b, and c you end up with the simple equation that tan pi = 0, hence

a + b + c - abc = 0
hence
a + b + c = abc

try that method. just for those of you who awanted a more systematic proof and that made more induction sort of sense. thanks again guys.