Question

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

Answer 1

`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`

`[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`

Answer 2

see below

Explanation:

see attached for my workings, step-by-step and the trig identities I used.