Prove that `sum_(r=0)^(n) ""^(n)C_(r )sin rx. cos (n-r)x = 2^(n-1) xx sin nx`.

Explanation

`S = underset(r=0)overset(n)sum.^(n)C_(r)sin rx . Cos (n-r) x`
` = .^(n)C_(0) sin 0x cos n x + .^(n)C_(1) sin x cos (n-1)x`
`+ .^(n)C_(2)sin 2x cos (n-2)x"….." + .^(n)C_(n-1)sin (n-1) x cos x`
`+ .^(n)C_(n)sin nx cos 0x`
Writing the sum in reverse order, we get
`S = .^(n)C_(n) sin nx cos 0x + .^(n)C_(n-1) sin(n-1)xcos x`
`+ .^(n)C_(n-2)sin(n-2)x cos 2x + "......."`
`+ .^(n)C_(1) sinx cos (n-1)x+.^(n)C_(0) sin 0x cos nx`
`:. S = .^(n)C_(0) sin nx cos 0x + .^(n)C_(1) sin (n-1)x cos x`
`+ .^(n)C_(2) sin(n-2)x cos 2x + "......"`
`+ .^(n)C_(n-1) sinx cos(n-1)x + .^(n)C_(n) sin 0x cos x nx`
Adding (1) and (2), we get
`2S = .^(n)sin(0x+nx) + .^(n)C_(1)sin(x+(n-1)x)`
`+ .^(n)C_(2) sin (2x+(n-2)x) + "......" + .^(n)C_(n) sin (nx+0x)`
` = (.^(n)C_(0) + .^(n)C_(1)+.^(n)C_(2)+.^(n)C_(3)+"......"+.^(n)C_(n)) sin nx`
`= 2^(n) sin nx`
`:. S = 2^(n-1) sin nx`