INDUCTION

Let P(n) = sinx +sin2x..sin(nx)
when n =1,
sinx = sin(1+1)x/2*sinx/2sinx/2
Hence P(n) is true

Assuming P(n) is true,
P(n+1)= sinx + sin2x+...sin(nx)+sin(n+1)x
= sin(n+1)x/2*sin(nx)sinx/2 + sin(n+1)x

[sin(n+1)x = 2sin(n+1)x/2 * cos(n+1)x/2

therefore p(n+1)=
sin(n+1)x2(sin(nx)+2sinx/2 * cos(n+1)x/2sinx/2)

=sin(n+1)x2(sin(nx)/2-sin(nx)/2+sin(n+2)x/2sin(x/2))

[BECAUSE 2cosA*sinB = sin(A+B)-sin(A-B)]

=sin(n+1)x/2 * sin(n+2)x/2sin(x/2)

Which gives us the required result.