Question

cosC-cosD=2sin(C+D/2) .sin(D-C/2)
with proof​

Answer 1

$\large\tt{\red{Let\:LHS=cosC-cosD}}$

$\large\tt{\red{\:\:\:\:\:\:\:\:RHS=2sin(\frac{C+D}{2})sin(\frac{D-C}{2})}}$

$\large\tt{\pink{Let\:(C=A+B)\:and\:(D=A-B)}}$

$\large\tt{\pink{A=\frac{C+D}{2}\:and\:B=\frac{C-D}{2}}}$

$\large\tt{\green{LHS=cosC-cosD}}$

$\large\sf{\green{=cos(A+B)-cos(A-B)}}$

$\large\sf{\green{=cosA.cosB-sinA.sinB}}$

$\large\sf{\green{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-(cosA.cosB+sinA.sinB)}}$

$\large\sf{\green{=cosA.cosB-sinA.sinB}}$

$\large\sf{\green{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-cosA.cosB-sinA.sinB}}$

$\large\sf{\green{=-sinA.sinB-sinA.sinB}}$

$\large\sf{\green{=-2sinA.sinB}}$

$\large\sf{\green{=-2sin(\frac{C+D}{2}).sin(\frac{C-D}{2})}}$

$\large\tt{\green{=2sin(\frac{C+D}{2}).sin(\frac{D-C}{2})=RHS}}$