Forever Sine

Using the given equation $\sin^{2}(\theta) + \sin(\theta) = 1$ we have that

$\sum_{n=1}^{6} \sin^{n}(\theta) = (\sin^{2}(\theta) + \sin(\theta))(\sin^{4}(\theta) + \sin^{2}(\theta) + 1) =$

$\sin^{4}(\theta) + \sin^{2}(\theta) + 1 = \sin^{4}(\theta) + (1 - \sin^{2}(\theta))^{2} + 1 =$

$2 - 2\sin^{2}(\theta)(1 - \sin^{2}(\theta)) = \boxed{2 - 2\sin^{2}(\theta)\cos^{2}(\theta)}.$

Given that,

$\large sin^2 \theta + sin \theta = 1$

$\large \Rightarrow sin \theta = 1 - sin^2 \theta = cos^2 \theta$

Now.,

$\large sin^{6} \theta + sin^{5} \theta + sin^{4} \theta + sin^{3} \theta + sin^{2} \theta + sin \theta$

$= \large sin^6 \theta + sin^5 \theta + sin^4 \theta + cos^6 \theta + sin^2 \theta + cos^2 \theta$

$= ( \large sin^2 \theta + cos^2 \theta ) ( sin^4 \theta - sin^2 \theta cos^2 \theta + cos^4 \theta ) + sin^5 \theta + sin^4 \theta + 1$

$= \large sin^4 \theta - sin^3 \theta + cos^4 \theta + sin^5 \theta + sin^4 \theta + 1$

$= ( \large sin^2 \theta + cos^2 \theta ) - 3 sin^3 \theta + sin^5 \theta + sin^4 \theta + 1$

$=\large sin^4 \theta + sin^3 \theta ( sin^2 \theta - 3 ) + 2$

$= \large sin^4 \theta + sin^3 \theta ( - 2 - cos^2 \theta ) + 2$

$= \large sin^4 \theta - sin^4 \theta - 2 cos^2 \theta sin^2 \theta + 2$

$= \large 2 - 2 sin^2 \theta cos^2 \theta$