Hey just a trig question
cos(A/2)+cos(B/2)+cos(C/2)=4cos[(pi-A)/4]cos[(pi-B)/4]cos[(pi-C)/4]
Sorry about the horrible typing. How do I prove this? Or just give me a hint on what trig identity to use. Thanks!
CO

Question

Hey just a trig question
cos(A/2)+cos(B/2)+cos(C/2)=4cos[(pi-A)/4]cos[(pi-B)/4]cos[(pi-C)/4]
Sorry about the horrible typing. How do I prove this? Or just give me a hint on what trig identity to use. Thanks!
CO

crashonce · Answer

If you can't be bothered to solve it etc. just tell me which identity to use. That's where i'm stuck

shadowlegendx. · Answer

Not many users are on that can assist with such a question

shadowlegendx. · Answer

It's summertime for me and I hate trig, too lazy for such a question, haha sorry :P

crashonce · Answer

its ok lol

mathmate · Answer

Use half angle formula, and addition formula.

mathmate · Answer

Anyway, cos(pi-A)=-cosA.|dw:1406112719172:dw|

crashonce · Answer

I was thinking I could substitute A/2 as D and B/2 as E etc. would that work?

mathmate · Answer

Yes, but you might have a problem on the right hand side. I would go for the half angle formula, such as: http://www.sosmath.com/trig/douangl/douangl.html

crashonce · Answer

ahh right thanks!

mathmate · Answer

@CrashOnce
I have an impression that there is an additional condition that A+B+C=$\pi$
Would you like to recheck the question?

crashonce · Answer

yes there is sorry!

shadowlegendx. · Answer

@shamil98 just cuz I can