Question 114238
Euler's Formula is:


{{{e^(ix)=cos(x)+i*sin(x)}}}


Apply this formula to {{{x=alpha+beta}}}


{{{e^(i(alpha+beta))=cos(alpha+beta)+i*sin(alpha+beta)}}}


But {{{e^(i(alpha+beta))=e^(i*alpha)e^(i*beta)}}}


So {{{e^(i(alpha+beta))=(cos(alpha)+i*sin(alpha))(cos(beta)+i*sin(beta))}}}


Which is to say, {{{cos(alpha+beta)+i*sin(alpha+beta)=(cos(alpha)+i*sin(alpha))(cos(beta)+i*sin(beta))}}}


Expanding the right side using FOIL yields:
{{{cos(alpha+beta)+i*sin(alpha+beta)=cos(alpha)cos(beta)+cos(alpha)*i*sin(beta)+i*sin(alpha)cos(beta)+(i^2)sin(alpha)sin(beta)}}}


And gathering like terms:
{{{cos(alpha+beta)+i*sin(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)+i(cos(alpha)sin(beta)+sin(alpha)cos(beta))}}}


Since we know that {{{a+bi=c+di}}} if and only if {{{a=c}}} and {{{b=d}}},
we can see that:


{{{cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)}}}, which is the desired identity.


By the way, we have also proven the companion identity:


{{{sin(alpha+beta)=cos(alpha)sin(beta)+sin(alpha)cos(beta)}}}


Hope this helps,
John