Prove the identity: 2sin(a+b)sin(a-b) = cos(2b)-cos(2a)

2 answers:

4 0

$2sin(a+b)sin(a-b)=cos(2b)-cos(2a) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{let's do this part first, we'll add the "2" later}}{sin(a+b)sin(a-b)} \\\\\\ \underset{\textit{difference of squares}}{[sin(a)cos(b)+cos(a)sin(b)][sin(a)cos(b)-cos(a)sin(b)]} \\\\\\ sin^2(a)cos^2(b)-cos^2(a)sin^2(b) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{we know that}}{cos(2\theta )=cos^2(\theta)-sin^2(\theta)}\implies sin^2(\theta)=cos^2(\theta)-cos(2\theta ) \\\\[-0.35em] \rule{34em}{0.25pt}$

$[cos^2(a)-cos(2a)]cos^2(b)-cos^2(a)[cos^2(b)-cos(2b)] \\\\\\ \underline{cos^2(a)cos^2(b)}-cos(2a)cos^2(b)\underline{-cos^2(a)cos^2(b)}+cos^2(a)cos(2b) \\\\\\ cos^2(a)cos(2b)-cos(2a)cos^2(b) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{we also know that}}{cos(2\theta)=2cos^2(\theta)-1}\implies \cfrac{cos(2\theta)+1}{2}=cos^2(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}$

$\left[ \cfrac{cos(2a)+1}{2} \right]cos(2b)-cos(2a)\left[ \cfrac{cos(2b)+1}{2} \right] \\\\\\ \stackrel{\textit{now let's bring back the "2"}}{2\left( \left[ \cfrac{cos(2a)+1}{2} \right]cos(2b)-cos(2a)\left[ \cfrac{cos(2b)+1}{2} \right] \right)} \\\\\\\ [cos(2a)+1]cos(2b)-cos(2a)[cos(2b)+1] \\\\\\ \underline{cos(2a)cos(2b)}+cos(2b)-\underline{cos(2a)cos(2b)}-cos(2a)\implies cos(2b)-cos(2a)$