how would you use induction to prove this:
$\sin(x)-sin(3x)+sin(5x)-...+(-1)^{(n+1)}sin[(2n-1)x] = \frac{(-1)^{(n+1)}sin2nx}{2cosx} $
I know how you assume its true for n=k, and then prove for n=k+1, but I get to
Left Hand Side: $\frac{(-1)^{(k+1)}sin2kx}{2cosx}+(-1)^{k+2}sin[(2k+1)x]$ but I'm not sure what step to take next.
any help would be appreciated. Cheers
We can write the LHS of the inductive step as:
$$(-1)^k (-1) \frac {\sin 2kx}{2\cos x} + (-1)^k (-1)^2 [\sin 2kx \cos x + \cos 2kx \sin x] $$ (remember the formula for $\sin (\alpha + \beta) = \sin \alpha \cos \beta +\cos \alpha \sin \beta$? Here $\alpha = 2kx$ and $\beta = x $) $$ = (-)^k [ \sin 2kx \cos x + \cos 2kx \sin x - \frac {\sin 2kx}{2 \cos x}]$$ $$=(-1)^k [ \frac {\sin 2kx \cos 2x}{2\cos x} + \cos 2kx \sin x]$$ $$=(-1)^k [ \frac {\sin 2kx \cos 2x}{2 \cos x} + \frac {\cos 2kx \sin 2x}{2 \cos x}]$$ $$=(-1)^{k+2} \frac {\sin 2 (k+1)x}{2\cos x} $$ and we are done! Hope it helps.
If $\sum_{r=1}^m(-1)^r\sin(2r-1)x=\dfrac{(-1)^{m+1}\sin2mx}{2\cos x}$
$$\sum_{r=1}^{m+1}(-1)^r\sin(2r-1)x=\sum_{r=1}^m(-1)^r\sin(2r-1)x++(-1)^{m+2}\sin(2m+1)x=\dfrac{(-1)^{m+1}\sin2mx}{2\cos x}+(-1)^{m+2}\sin(2m+1)x$$
$$=\dfrac{(-1)^{m+1}}{2\cos x}[\sin2mx-2\sin(2m+1)x\cdot\cos x]$$
Now $2\sin(2m+1)x\cdot\cos x=\sin2(m+1)x+\sin2mx$