# half angle conversion formulas

• September 26th 2013, 07:00 PM
fireboy25
half angle conversion formulas
sinA+sinB+sinC=4cos1/2Acos1/2Bcos1/2C

..guys can you help me in proving this one..tnx..
• September 26th 2013, 07:26 PM
Prove It
Re: half angle conversion formulas
Yikes this is hard to read, some brackets or spaces would be nice ><

Is it \displaystyle \begin{align*} \sin{(A)} + \sin{(B)} + \sin{(C)} = 4\cos{\left( \frac{A}{2} \right)} \cos{\left( \frac{B}{2}\right)}\cos{\left( \frac{C}{2}\right)} \end{align*}?
• September 26th 2013, 07:40 PM
topsquark
Re: half angle conversion formulas
Quote:

Originally Posted by fireboy25
sinA+sinB+sinC=4cos1/2Acos1/2Bcos1/2C

..guys can you help me in proving this one..tnx..

What have you been able to do with this so far?

-Dan
• September 26th 2013, 09:21 PM
ibdutt
Re: half angle conversion formulas
Also must write the condition for this question we should have A + B + C = 180 degree.
Then use the sum and product formula: sin P + sin Q = 2 sin ( P+Q)/2 cos ( P-Q)/2 etc.
• September 27th 2013, 01:05 PM
topsquark
Re: half angle conversion formulas
Quote:

Originally Posted by ibdutt
Also must write the condition for this question we should have A + B + C = 180 degree.

That's only if A, B, C are angles in a triangle. But it's a good question.

@fireboy25: Are A, B, and C the angles of a triangle?

-Dan
• September 27th 2013, 08:32 PM
ibdutt
Re: half angle conversion formulas
• September 29th 2013, 02:15 PM
fireboy25
Re: half angle conversion formulas
yes..the sum of A, B, C is 180..