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`
`rArr S=sinx cox 100x+sin2x cos 99x+..........+sin100x cos x`. (i)
`S=cos x sin 100x+cos 2x sin 99x+.......+sin x cos 100x`.....(ii)
(on writing in reverse order)
Adding Eqs. (i) and (ii) we get
`2S=(sin x cos 100x+cos x sin 100x)`
`+(sin2x cos 99x+cos 2xsin99x)`
`+(sin 100x cos x+sin x cos 100x)`
`=sin 101x+sin 101x+......+sin101x(100"times")`
Hence, `S=50sin(101x)`.