18. [11] Let $x$ and $y$ be positive real numbers such that $x^{2}+y^{2}=1$ and $\left(3 x-4 x^{3}\right)\left(3 y-4 y^{3}\right)=-\frac{1}{2}$. Compute $x+y$.
Answer: $\sqrt{\frac{\sqrt{6}}{2}}$ Solution 1: Let $x=\cos (\theta)$ and $y=\sin (\theta)$. Then, by the triple angle formulae, we have that $3 x-4 x^{3}=-\cos (3 \theta)$ and $3 y-4 y^{3}=\sin (3 \theta)$, so $-\sin (3 \theta) \cos (3 \theta)=-\frac{1}{2}$. We can write this as $2 \sin (3 \theta) \cos (3 \theta)=\sin (6 \theta)=1$, so $\theta=\frac{1}{6} \sin ^{-1}(1)=\frac{\pi}{12}$. Thus, $x+y=\cos \left(\frac{\pi}{12}\right)+\sin \left(\frac{\pi}{12}\right)=$ $\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{6}}{2}$.

Solution 2: Expanding gives $9 x y+16 x^{3} y^{3}-12 x y^{3}-12 x^{3} y=9(x y)+16(x y)^{3}-12(x y)\left(x^{2}+y^{2}\right)=-\frac{1}{2}$, and since $x^{2}+y^{2}=1$, this is $-3(x y)+16(x y)^{3}=-\frac{1}{2}$, giving $x y=-\frac{1}{2}, \frac{1}{4}$. However, since $x$ and $y$ are positive reals, we must have $x y=\frac{1}{4}$. Then, $x+y=\sqrt{x^{2}+y^{2}+2 x y}=\sqrt{1+2 \cdot \frac{1}{4}}=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}$.