Any idea on how WolframAlpha did this sum?

Question

So I've been trying to work out this sum for quite a while:

WolframAlpha unfortunately won't supply step by step proofs for this for some reason...

As for how i tried to prove this: I looked at sin((2n-1)*a) as the imaginary part of cis((2n-1)a), which is a geometric series (a_n = a_1q^(n-1), where a_1 = cis(a) and q=cis(2a)), therefore its sum from n=1 to n=k should be:

Meaning the sum is:

(Note: I used De Moivre's Theorem to put the n power inside the cis)

But no matter how i tried continuing from here i always arrived at a huge trigonometric expression which i did not see how it (the imaginary part) could become what WolframAlpha is saying it is...

Any directions/solutions would be appreciated. Thanks

Answer 1

(i) $$\sin a+ \sin 3a+...+\sin ((2k-1)a)$$ $$=\sin a+\sin 2a+...+\sin 2ka-(\sin 2a+\sin 4a+...+\sin 2ka)$$ (ii) Use forward differences $\Delta f(x)=f(x+1)-f(x)$ and $\sum_{k=1}^b \Delta f(k)=f(b+1)-f(1)$ $$\text{(iii) }\sin k \theta=(\Delta (\cos ((k-1/2)\theta)) )/(-2 \sin (\theta/2))$$ ()iv Put in (i) $\theta=a$ and $\theta =2a$ and subtract.