tl:dr at the end:
So I started of scribbling this simple function in my notebook
$ f(\theta) = (1+\sin\theta + \cos\theta)$
$ = \sin^2\theta + \cos^2\theta + \sin\theta + \cos\theta$
$ = (\sin\theta + \cos\theta)^2 + \sin\theta +\cos\theta - \sin2\theta$
$ = (\sin\theta + \cos\theta)*(1 + \sin\theta + \cos\theta) - \sin2\theta$
Doing this once again on the $1 + \sin\theta + \cos\theta $
$ = (\sin\theta + \cos\theta)*((\sin\theta + \cos\theta)^2 + (\sin\theta + \cos\theta)- \sin2\theta)) - \sin2\theta$
$ = (\sin\theta + \cos\theta)^2*(1 + \sin\theta + \cos\theta) - \sin2\theta(1+ \sin\theta + \cos\theta)$
after $n$ such steps:
$f(\theta) = (\sin\theta + \cos\theta)^n(1+\sin\theta + \cos\theta) - \sin2\theta*\Sigma_{k=0}^{k=n-1}(\sin\theta + \cos\theta)^k = 1 + \sin\theta + \cos\theta$
My question is how do I prove this identity true $\forall \space n \space \epsilon \space W$
$1 + \sin\theta + \cos\theta = (\sin\theta + \cos\theta)^n(1+\sin\theta + \cos\theta) - \sin2\theta*\Sigma_{k=0}^{k=n-1}(\sin\theta + \cos\theta)^k $
tl:dr
In a non inductive manner. I find it quite intriguing that the right hand side is independent of $n$ and I want to prove it without knowing it's true. Thank you.