Problem:
Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$. Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Solution:
We perform a recursion style solution on this problem, relying on the fact that $\sin^2x + \cos^2x = 1$Note that $$(\sin^2x+\cos^2x)(\cancel{\sin^2x+\cos^2x}) = \sin^4x + \cos^4x + 2 \sin^2x \cos^2x = 1 \implies \sin^4x + \cos^4x = 1 - 2 \sin^2x \cos^2x$$Then, $$(\sin^4x + \cos^4x)(\cancel{\sin^2x + \cos^2x}) = \sin^6x + \cos^6x + \sin^4x \cos^2x + \sin^2x \cos^4x = \sin^6x + \cos^6x + (\sin^2x \cos^2x)(\cancel{\sin^2x + \cos^2x}) = 1 - 2 \sin^2x \cos^2x \implies \sin^6x + \cos^6x = 1 - 3 \sin^2x \cos^2x$$Then, $$(\sin^6x + \cos^6x)(\cancel{\sin^2x + \cos^2x}) = \sin^8x + \cos^8x + \sin^6x \cos^2x + \sin^2x \cos^6x = \sin^8x + \cos^8x + (\sin^2x \cos^2x)(\sin^4x + \cos^4x) = \sin^8x + \cos^8x + (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x) = 1 - 3 \sin^2x \cos^2x \implies \sin^8x + \cos^8x = 1 - 3 \sin^2x \cos^2x - (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x)$$Then, $$(\sin^8x + \cos^8x)(\cancel{\sin^2x + \cos^2x}) = \sin^{10}x + \cos^{10}x + \sin^8x \cos^2x + \sin^2x \cos^8x = \sin^{10}x + \cos^{10}x + (\sin^2x \cos^2x)(\sin^6x + \cos^6x) = \sin^{10}x + \cos^{10}x + (\sin^2x \cos^2x)(1 - 3 \sin^2x \cos^2x) = 1 - 3 \sin^2x \cos^2x - (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x)$$At this point, we know that the value of $\sin^{10}x + \cos^{10}x = \frac{11}{36}$, so letting $y = \sin^2x \cos^2x$ and plugging these values into the above equation, we get $y=\frac{1}{6}$ and $y=\frac{5}{6}$, but we can weed out the second one because it gives a negative value for the computation of $\sin^4x + \cos^4x$, which is clearly not possible.Now, we apply the recursion one last time, to find the desired term.We have $$(\sin^{10}x+\cos^{10}x)(\cancel{\sin^2x+\cos^2x}) = \sin^{12}x+\cos^{12}x + (\sin^2x \cos^2x)(\sin^8x + \cos^8x) = \sin^{12}x+\cos^{12}x + (\sin^2x \cos^2x)(1 - 3 \sin^2x \cos^2x - (\sin^2x \cos^2x)(1 - 2 \sin^2x \cos^2x)) = \sin^{12}x + \cos^{12}x + (a)(1-3a-(a)(1-2a)) = \frac{11}{36}$$Solving this equation for $\sin^{12}x + \cos^{12}x$, we obtain $\boxed{\frac{13}{54}}$.
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