P.T sinx +sin3x+ . . . +sin(2n-1)x = [sin^(2) (nx)] / sinx for any natural number n using mathematical induction
So you must prove that $\displaystyle \displaystyle{\sum\limits^{n+1}_{k=1}\sin(2k-1)x=\frac{\sin^2(n+1)x}{\sin x}}$ , as the case $\displaystyle n=1$ is trivially true and assuming true for n:
$\displaystyle \displaystyle{\sum\limits^{n+1}_{k=1}\sin(2k-1)x=\sum\limits^n_{k=1}\sin(2k-1)x+\sin(2n+1)x=\frac{\sin^2nx}{\sin x}+\sin(2n+1)x}}=$
$\displaystyle =\displaystyle{\frac{\sin^2nx+\sin x\sin(2n+1)x}{\sin x}}$ . Thus, it's enough to show that $\displaystyle \sin^2(n+1)x=\sin^2nx+\sin x\sin(2n+1)x$.
For this you'll need identities: in the LHS use the well known $\displaystyle \sin(\alpha+\beta)=\sin\alpha\cos\beta+sin\beta\co s\alpha$ (watch
that square!), and on the RHS, second term, use the same identity.
After the above's done pass all the terms to the LHS, use the trigonometric Pythagoras identity
and also $\displaystyle \cos(2\beta)=1-2\sin^2\beta$ ...
Tonio