How do you prove that #sin(a+b)=sin a cos b + cos a sin b#?

1 Answer
George C.
Jan 13, 2018

See explanation...

Explanation:

Let us start from:

#e^(i theta) = cos theta + i sin theta#

Then:

#cos(a+b)+i sin(a+b)#

#= e^(i(a+b))#

#= e^(ia) * e^(ib)#

#= (cos(a)+i sin(a))(cos(b)+ i sin(b))#

#= (cos(a)cos(b)-sin(a)sin(b))+i(sin(a)cos(b)+ cos(a)sin(b))#

Equating real and imaginary parts we find:

#{ (cos(a+b) = cos(a)cos(b)-sin(a)sin(b)), (sin(a+b) = sin(a)cos(b)+cos(a)sin(b)) :}#

Matrix formulation

Equivalently, note that a matrix of the form:

#((cos theta, -sin theta), (sin theta, cos theta))#

represents a rotation by angle #theta# around the origin.

Then a rotation by angle #a+b# is equivalent to the combination of a rotation by angle #a# followed by a rotation by angle #b#.

So:

#( (cos(a+b), -sin(a+b)), (sin(a+b), cos(a+b)))#

#=((cos(b), -sin(b)),(sin(b), cos(b)))((cos(a), -sin(a)), (sin(a), cos(a)))#

#=((cos(a)cos(b)-sin(a)sin(b), -sin(a)cos(b)-cos(a)sin(b)), (sin(a)cos(b)+cos(a)sin(b), cos(a)cos(b)-sin(a)sin(b)))#

So equating matrix entries, we find:

#{ (cos(a+b) = cos(a)cos(b)-sin(a)sin(b)), (sin(a+b) = sin(a)cos(b)+cos(a)sin(b)) :}#

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