Use mathematical induction to prove

sin x + sin 3x+ sin 5x +..............+sin(2n-1)x = (sin nx)^2)/sin x

With n = 1, we see that this reduces to:
sin(x) = sin²x/sin(x), which is true.

Now, we need to show that if this statement is true for some integer k, then it is true for k + 1.

Since the statement holds for n = k:
sin(x) + sin(3x) + sin(5x) + ... + sin[(2k - 1)x] = sin²(kx)/sin(x).

Now, we need to show that:
sin(x) + sin(3x) + sin(5x) + ... + sin[(2k - 1)x] + sin{[2(k + 1) - 1]x} = sin²[(k + 1)x]/sin(x).

To do this, note that:
sin(x) + sin(3x) + sin(5x) + ... + sin[(2k - 1)x] + sin{[2(k + 1) - 1]x}
= sin²(kx)/sin(x) + sin{[2(k + 1) - 1]x}, by the inductive hypothesis
= sin²(kx)/sin(x) + sin[(2k + 1)x]
= {sin²(kx) + sin[(2k + 1)x]sin(x)}/sin(x), by combining fractions.

We can write,
sin (2k+1)x = sin [(k+1) + k]x = sin (k+1) cos kx - cos (k+1)x sin kx
Then we have
=sin²(kx) + sin[(2k + 1)x]sin(x) = sin² kx + [sin (k+1) cos kx - cos (k+1)x sin kx]sinx

= sin (k+1)x cos kx sin x + sin (k + 1)x cos x sin kx - sin (k + 1)x cos x sin kx
+ cos(k+1)x sinkx sinx + sin² k x
= sin (k+1)x [ sinx cos kx + cosx sin kx] + sinkx [ cos (k+1)x sin x - sin (k+1)x cosx] + sin² k x

= sin (k+1)x sin (k+1)x + sinkx sin[ x - (k+1)x] + sin² k x

= sin² (k + 1)x + sinkx sin (-kx) + sin² k x

= sin² (k + 1)x + sinkx(- sin kx) + sin² k x

= sin² (k + 1)x
Using this:
{sin²(kx) + sin[(2k + 1)x]sin(x)}/sin(x) = sin²[(k + 1)x]/sinx

so, by the Principle of Mathematical Induction:
sin(x) + sin(3x) + sin(5x) + ... + sin[(2n - 1)x] = sin²(nx)/sin(x), as required.