# Thread: soluiton required up to n terms

1. ## soluiton required up to n terms

Find Sina - Sin2a + Sin3a - Sin4a ………. n terms.

2. First of all we have two cases: when n is even or n is odd.
I'll do the case when n is even. I think you can do the other case.
$S=\sin\alpha-\sin 2\alpha+\sin 3\alpha-\sin 4\alpha +\ldots+\sin(2n-1)\alpha-\sin 2n\alpha$
Split the sum in two other sums:
$\displaystyle S_1=\sin\alpha+\sin 3\alpha+\ldots+\sin(2n-1)\alpha=\frac{\sin^2n\alpha}{\sin\alpha}$
$\displaystyle S_2=\sin 2\alpha+\sin 4\alpha+\ldots+\sin 2n\alpha=\frac{\sin n\alpha\sin(n+1)\alpha}{\sin\alpha}$
To calculate each sum, multiply both sides by $\sin\alpha$ and transform the products of sines in differences of cosines.
Now $\displaystyle S=S_1-S_2=\frac{\sin n\alpha(\sin n\alpha-\sin(n+1)\alpha)}{\sin\alpha}\displaystyle=-\frac{\sin n\alpha\cos\frac{(2n+1)\alpha}{2}}{\cos\frac{\alph a}{2}}$

3. Use complex numbers. And the fact that $i\sin x = \sinh (ix) = (e^{ix}-e^{-ix})/2$. Now use geometric series.