Question

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

Answer 1

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)`
`:.sum_(k=1)^100 sin(kx)cos(101-k)x = 50sin(101x)`