Solving $45x-3795x^3 +95634x^5 - \cdots + 945x^{41}-45x^{43}+x^{45} = N$? We have the following excerpt from John Stillwell's Mathematics and its History:



My question is, what are the hints which give away the fact that this comes from the expansion of $\sin 45 \theta$?

Perhaps, it is this: 
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}\\= \color{blue}{\displaystyle\sum_{r=0, 2r+1\le n}(-1)^r\dbinom{n}{2r+1}\cos^{n-2r-1}\theta \sin^{2r+1}\theta} $
Proof: 
$(\cos \theta+ i\sin \theta)^n = \cos n\theta + i\sin n\theta$
Writing the binomial expansion of LHS we get: 
$\dbinom{n}{0}\cos^n \theta + i\dbinom{n}{1}\cos^{n-1}\theta\sin \theta -\dbinom{n}{2}\cos^{n-2}\theta \sin^2\theta - i\dbinom {n}{3}\cos^{n-3}\theta \sin \theta+\dbinom{n}{4}\cos^{n-4}\theta\sin ^4 \theta+ i\dbinom{n}{5}\cos^{n-5}\theta\sin \theta$
Comparing real and imaginary parts, 
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}$... which is the desired result. 
