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sin a +sin b +sin c=cos a+cos b+cos c then prove #cos 2a+cos 2b+cos 2c=0 #sin 2a+sin 2b+sin 2c=0 #cos 3a+cos 3b+cos 3c= 3cos(a+b+c)

sin a +sin b +sin c=cos a+cos b+cos c then prove
#cos 2a+cos 2b+cos 2c=0
#sin 2a+sin 2b+sin 2c=0
#cos 3a+cos 3b+cos 3c=
3cos(a+b+c)
 

Grade:12th pass

2 Answers

girish
14 Points
8 years ago
you left part of the question cosa+cosb+cosc=0=sina+sinb+sinc you can prove using complex numbers. we can use that 2cosn(argz)=z^n+1/z^n
Gowri sankar
292 Points
8 years ago
The question is cosa+cosb+cosc=sina+sinb+sinc=0 this is proov using in complex numbers.use the formula 2cosn(argz)=zn+1/zn.
 

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