<s>[INST] If $a=\sin\theta+\cos\theta$ and $b=\cos\theta-\sin\theta$, express $\sum_k\sin^k\theta(\sin^k\theta-\cos^k\theta)$ in terms of $a$ and $b$ 
If $a=\sin\theta+\cos\theta$ and $b=\cos\theta-\sin\theta$, then express
  $$\sin \theta(\sin \theta-\cos \theta)+\sin^2\theta(\sin^2\theta-\cos^2\theta)+\sin^3\theta(\sin^3\theta-\cos^3\theta) +\cdots$$
  in terms of $a$ and $b$.

My observation :
$\underbrace{\sin \theta(\sin \theta-\cos \theta)}_1 +$$\underbrace{\sin^2\theta(\sin^2\theta-\cos^2\theta)}_2 +\underbrace{\sin^3\theta(\sin^3\theta-\cos^3\theta)}_3 +\cdots\\$ 
$i.)$$\;\sin \theta(\sin \theta-\cos \theta) =\sin \theta (-b) =\bbox[4px, yellow] {-b\sin \theta} $ $\\$
$ii.) \;\sin^2 \theta(\sin^2 \theta-\cos^2 \theta)=\sin^2 \theta(\sin\theta-\cos
 \theta)(\sin \theta + \cos\theta)=\bbox[yellow, 4px]{-ab\sin^2\theta}
\\$
$\mathrm {Now} \\$
$iii.)$ $\sin^3\theta(\sin^3\theta-\cos^3\theta)=\sin^3\theta (\sin\theta-\cos\theta)(\sin^2\theta +\sin\theta \cos\theta+\cos^2\theta) \\$
$=\sin^3\theta(-b)\left(1+\boxed {\sin\theta\cos\theta} \rightarrow\left\{\frac{a^2-b^2} {4} \right\} \right) \\$
$=\underline{\bbox[yellow, 4px] {-b\sin^3\theta\left(1+\frac{a^2-b^2}{4}\right)}}$
Can anyone suggest a way out? 
 [/INST] Subtracting b from a we have $a-b=\sin(x)+\cos(x)-(\cos(x)-\sin(x))$ thus $\frac{a-b}{2}=\sin(x)$ 
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