• May 10th 2006, 08:26 PM
mpgc_ac
:confused: Given a triangle ABC let
X=tan[(B-C)/2]tan(A/2)
Y=tan[(C-A)/2]tan(B/2)
Z=tan[(A-B)/2]tan(C/2)
Prove that X+Y+Z+XYZ=0
• May 27th 2006, 03:15 AM
chancey
Quote:

Originally Posted by mpgc_ac
:confused: Given a triangle ABC let
X=tan[(B-C)/2]tan(A/2)
Y=tan[(C-A)/2]tan(B/2)
Z=tan[(A-B)/2]tan(C/2)
Prove that X+Y+Z+XYZ=0

Some formulae that might help:
If $t = \tan \frac{\theta}{2}$
Then $\tan \theta = \frac{2 t}{1 - t^2}$
And:
$tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1+\tan \theta \tan \phi}$

I tried, but the algebra is too much, there must be another way
• May 30th 2006, 12:58 AM
malaygoel
X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
Similar are Y and Z
Then solve (X+Y)/XY+1
You will find it is equal to –Z
Your result immediately follows
You can prove Napier Analogy by using Sine Law
• May 30th 2006, 04:46 AM
chancey
Quote:

Originally Posted by malaygoel
X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
Similar are Y and Z
Then solve (X+Y)/XY+1
You will find it is equal to –Z
Your result immediately follows
You can prove Napier Analogy by using Sine Law

Can you show a bit more detail? I havn't seen this method before...
• May 30th 2006, 07:07 AM
malaygoel
Quote:

Originally Posted by chancey
Can you show a bit more detail? I havn't seen this method before...

Have you tried?
• May 30th 2006, 07:14 AM
malaygoel
Quote:

Originally Posted by malaygoel
X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
Similar are Y and Z
Then solve (X+Y)/XY+1
You will find it is equal to –Z
Your result immediately follows
You can prove Napier Analogy by using Sine Law

X=(b-c)/(b+c)
Y=(c-a)/(c+a)
(X+Y)/(XY+1)={(b-c)/(b+c)+(c-a)/(c+a)}/[{(c-a)/(c+a)}*{(b-c)/(b+c)}+1]
=(b-a)/(b+a)=-Z
therefore,
(X+Y)/(XY+1)=-Z
multiply both sides by XY+1
X+Y=-Z-XYZ
X+Y+Z+XYZ=0
Good Luck :)
• May 30th 2006, 07:20 AM
malaygoel
Proof of NApier Analogy
Quote:

Originally Posted by malaygoel
X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
You can prove Napier Analogy by using Sine Law

sine law states that
a/sinA =b/sinB =c/sinC =2R
(b-c)/(b+c)=(2RsinB-2RsinC)/(2RsinB+2RsinC)
=(sinB-sinC)/(sinB+sinC)
I can't write next step, use addition and subtraction fromulae and you will get X
• May 31st 2006, 02:55 AM
chancey
Quote:

Originally Posted by malaygoel
sine law states that
a/sinA =b/sinB =c/sinC =2R
(b-c)/(b+c)=(2RsinB-2RsinC)/(2RsinB+2RsinC)
=(sinB-sinC)/(sinB+sinC)
I can't write next step, use addition and subtraction fromulae and you will get X

Ahhh... I sort of get it, and I learnt a new sine rule, cool thanks.
• Jun 1st 2006, 05:21 AM
malaygoel
Quote:

Originally Posted by chancey
Ahhh... I sort of get it, and I learnt a new sine rule, cool thanks.

COULD YOU PROVE THE SINE RULE? :)
• Jun 1st 2006, 08:37 AM
ThePerfectHacker
Quote:

Originally Posted by malaygoel
COULD YOU PROVE THE SINE RULE? :)

In fact we can do better.
The extended sines law