if $\cos^{10}x + \sin^{10}x=11/36$, find $\cos^{12}x+\sin^{12}x$ If $\cos^{10}x+ \sin^{10}x=11/36$, find $\cos^{12}x+\sin^{12}x$. I've tried solving using index manipulation but no way. I think we're to use $\sin^2 x+\cos^2 x=1$. But I don't know how. Need help please.

Here is one solution. Let $A=\sin^2 x\cos^2 x$
$$\sin^4+\cos^4=\sin^2+\cos^2-2\sin^2 x\cos^2 x=1-2A$$
$$\sin^6+\cos^6=\sin^4+\cos^4-A(\sin^2 +\cos^2 x)=1-3A$$
$$\sin^8+\cos^8=\sin^6+\cos^6-A(\sin^4 +\cos^4 x)=1-4A+2A^2$$
$$\sin^{10}+\cos^{10}=\sin^8+\cos^8-A(\sin^6 +\cos^6 x)$$
$$=1-5A+5A^2$$
Thus we have $$5A^2-5A+1=\frac{11}{36}$$
Which gives $$A=\frac{1}{6}$$ using the fact that $A\leq \frac{1}{4}$
So finally
$$\sin^{12}+\cos^{12}=\sin^{10}+\cos^{10}-A(\sin^8 +\cos^8 x)$$
$$=1-6A+9A^2-2A^3=\frac{13}{54}$$
Hope I havent made a mistake.
