11-06-11 Prove sin(a+b) expansion...

Btw, just to satiate my curiosity.. Whats the reward sir?[3] My chocolate is still pending! [2]

sir i think maybe this can work...

consider a right triangle ABC
right angled at C

we drop a perpendicular from C on AB at D
clearly
LACD = B
NOW,
IN TRI ADC
SIN A = CD/AC , COS B = CD/AC
COS A = AD/AC, SIN B = AD/AC
THUS
SINA.COSB + COSA.SINB = 1

ALSO, A+B =90
SO, SIN(A+B) = 1

SO WE CAN CONCLUDE the exp of

SIN(A+B)

Proof by calculus:

checkout post #6,#7,#8 here

http://www.targetiit.com/iit-jee-forum/posts/function-18569.html

How about a complex numbers proof?

e^{i(a + b)} = cos(a + b) + isin(a + b).

But e^{i(a + b)} = e^{ia + ib} = e^iae^ib
= [cosa + isina][cosb + isinb]
= cosa.cosb + icosa.sinb + isina.cosb - sina.cosb

But this equals cos(a + b) + isin(a + b).
Equating imaginary parts,
sin(a + b) = sina.cosb + cosa.sinb

Note : This popped out of my head from nowhere...now I wish I'd used my head as much back in 11th and 12th..