How will you prove the formula #sin(A+B)=sinAcosB+cosAsinB# using formula of scalar product of two vectors?

1 Answer
Cesareo R.
Oct 2, 2016

See below.

Explanation:

Given

#vec v_(alpha) = (sin alpha, cos alpha)# and
#vec v_(beta) = (sin beta, cos beta)#

both unit vectors, because

#norm( vec v_(alpha)) = norm( vec v_(beta)) =1#

Have their inner product given by

#<< vec v_(alpha), vec v_(beta) >> = sin alpha sin beta+cos alpha cos beta# and which is the projection on #vec v_(alpha)# onto #vec v_(beta)# or vice-versa.

This is exactly #cos(alpha-beta) = cos(beta-alpha)#

If we take complex numbers and using the de Moivre's identity which reads

#e^(ix) = cos x + i sin x# we have

#(cos alpha + i sin alpha)(cos beta + i sin beta) = e^(ialpha)e^(ibeta)# #=cosalpha cosbeta-sinalpha sinbeta+i(sinalpha cosbeta+cosalpha sinbeta) =e^(i(alpha+beta)) = cos(alpha+beta)+isin(alpha+beta)#.

In the case of vectors we have projections and in the case of complex numbers we have rotations.

Using vectors with additional phase additions we can obtain also

#sin alpha cos beta + cosalpha sinbeta=sin(alpha+beta)#

It is left as an exercise.

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