Prove that: `sum_(k=1)^(100)sin(k x)cos(101-k)x=50"sin"(101 x)`

Explanation

Let `S = sum_(k=1)^100 sin(kx)cos(101-k)x`

`=>S = sinxcos100x+sin2xcos99x+...sin100xcosx->(1)`

Writing it in reverse order,

`=>S = sin100xcosx+sin99xcos2x+...sinxcos100x ->(2)`

Adding (1) and (2),

`=>2S = [(sinxcos100x+sin100xcosx)+(sin2xcos99x+sin99xcos2x)+...(sin100xcosx+sinxcos100x)]`

`=>2S = 100sin(x+100x)`

`=>S = 50sin(101x)`

...