2019 AIME Problem 8

Problem:

Let $x$ be a real number such that ${\sin }^{10}x+{\cos }^{10}x=\frac{11}{36}$. Then ${\sin }^{12}x+{\cos }^{12}x=\frac{m}{n}$ 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 ${\sin }^{2}x+{\cos }^{2}x=1$Note that $$({\sin }^{2}x+{\cos }^{2}x)(\cancel{{\sin }^{2}x+{\cos }^{2}x})={\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x=1⟹{\sin }^{4}x+{\cos }^{4}x=1-2{\sin }^{2}x{\cos }^{2}x$$Then, $$({\sin }^{4}x+{\cos }^{4}x)(\cancel{{\sin }^{2}x+{\cos }^{2}x})={\sin }^{6}x+{\cos }^{6}x+{\sin }^{4}x{\cos }^{2}x+{\sin }^{2}x{\cos }^{4}x={\sin }^{6}x+{\cos }^{6}x+({\sin }^{2}x{\cos }^{2}x)(\cancel{{\sin }^{2}x+{\cos }^{2}x})=1-2{\sin }^{2}x{\cos }^{2}x⟹{\sin }^{6}x+{\cos }^{6}x=1-3{\sin }^{2}x{\cos }^{2}x$$Then, $$({\sin }^{6}x+{\cos }^{6}x)(\cancel{{\sin }^{2}x+{\cos }^{2}x})={\sin }^{8}x+{\cos }^{8}x+{\sin }^{6}x{\cos }^{2}x+{\sin }^{2}x{\cos }^{6}x={\sin }^{8}x+{\cos }^{8}x+({\sin }^{2}x{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x)={\sin }^{8}x+{\cos }^{8}x+({\sin }^{2}x{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x)=1-3{\sin }^{2}x{\cos }^{2}x⟹{\sin }^{8}x+{\cos }^{8}x=1-3{\sin }^{2}x{\cos }^{2}x-({\sin }^{2}x{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x)$$Then, $$({\sin }^{8}x+{\cos }^{8}x)(\cancel{{\sin }^{2}x+{\cos }^{2}x})={\sin }^{10}x+{\cos }^{10}x+{\sin }^{8}x{\cos }^{2}x+{\sin }^{2}x{\cos }^{8}x={\sin }^{10}x+{\cos }^{10}x+({\sin }^{2}x{\cos }^{2}x)({\sin }^{6}x+{\cos }^{6}x)={\sin }^{10}x+{\cos }^{10}x+({\sin }^{2}x{\cos }^{2}x)(1-3{\sin }^{2}x{\cos }^{2}x)=1-3{\sin }^{2}x{\cos }^{2}x-({\sin }^{2}x{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x)$$At this point, we know that the value of ${\sin }^{10}x+{\cos }^{10}x=\frac{11}{36}$, so letting $y={\sin }^{2}x{\cos }^{2}x$ 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 ${\sin }^{4}x+{\cos }^{4}x$, which is clearly not possible.Now, we apply the recursion one last time, to find the desired term.We have $$({\sin }^{10}x+{\cos }^{10}x)(\cancel{{\sin }^{2}x+{\cos }^{2}x})={\sin }^{12}x+{\cos }^{12}x+({\sin }^{2}x{\cos }^{2}x)({\sin }^{8}x+{\cos }^{8}x)={\sin }^{12}x+{\cos }^{12}x+({\sin }^{2}x{\cos }^{2}x)(1-3{\sin }^{2}x{\cos }^{2}x-({\sin }^{2}x{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x))={\sin }^{12}x+{\cos }^{12}x+(a)(1-3a-(a)(1-2a))=\frac{11}{36}$$Solving this equation for ${\sin }^{12}x+{\cos }^{12}x$, we obtain $\boxed{{\displaystyle \frac{13}{54}}}$.