Trigonometry! #93

$a^2+b^2=(a+b)^2-2ab,\;a^3+b^3=(a+b)^3-3ab(a+b)$

Let $\sin^2\theta=a,\;\cos^2\theta=b$ and knowing that $a+b=1$ the question simplifies to:

$2[(a+b)^3-3ab(a+b)]-3[(a+b)^2-2ab]=2(1-3ab)-3(1-2ab)$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;=\boxed{-1}$

just put theta=0

Since $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$ we have that

$\sin^{6}(\theta) + \cos^{6}(\theta) = (\sin^{2}(\theta) + \cos^{2}(\theta)(\sin^{4}(\theta) - \sin^{2}(\theta)\cos^{2}(\theta) + \cos^{4}(\theta)) =$

$\sin^{4}(\theta) - \sin^{2}(\theta)\cos^{2}(\theta) + \cos^{4}(\theta)$ ,

since $\sin^{2}(\theta) + \cos^{2}(\theta) = 1.$ The given expression then simplifies to

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

My solution is same as of Brian Charlesworth.