If $2(\sin A-\sin^3 A)=\cos B$ and $2(\cos A+\cos^3 A)=\sin B$; $0<A,B<\pi/2$, then $\cos B=\sqrt{\frac{m}{n}}$ where $m,n$ are co-prime and $m+n$ is
Step-by-Step Solution
Key Concept: Square and add: $4(\sin A-\sin^3 A)^2+4(\cos A+\cos^3 A)^2=1$; simplify using $\sin^2+\cos^2=1$.
Squaring and adding: $4[\sin^2 A(1-\sin^2 A)^2+\cos^2 A(1+\cos^2 A)^2]=1$. Let $c=\cos^2 A$: $4[(1-c)c^2+(c)(1+c)^2]$... wait let $s=\sin^2 A$: $4[s(1-s)^2+(1-s)(2-s)^2]$... Numerically try $A=\pi/6$: $\sin A=1/2,\cos A=\sqrt{3}/2$. $2(1/2-1/8)=2(3/8)=3/4=\cos B$. $2(\sqrt{3}/2+3\sqrt{3}/8)=2\cdot 7\sqrt{3}/8=7\sqrt{3}/4=\sin B$. Check: $(3/4)^2+(7\sqrt{3}/4)^2=9/16+147/16=156/16\neq 1$. Try systematic: LHS sum of squares $=\cos^2 B+\sin^2 B=1$. So $4[s(1-s)^2+(1-s)(1+1-s)^2]=1$. Let $s=\sin^2 A$: $4[s(1-s)^2+(1-s)(2-s)^2]=1$. Let $t=1-s$: $4[(1-t)t^2+t(1+t)^2]=4t[t(1-t)+(1+t)^2]=4t[t-t^2+1+2t+t^2]=4t[1+3t]=4t+12t^2=1$. $12t^2+4t-1=0$. $t=\frac{-4\pm\sqrt{16+48}}{24}=\frac{-4\pm 8}{24}$. $t=1/6$ (positive). $\cos^2 A=1-s=t=1/6$. $\cos B=2\sin A(1-\sin^2 A)=2\sin A\cdot t$... $\sin^2 A=1-t=5/6$, $\sin A=\sqrt{5/6}$. $\cos B=2\sqrt{5/6}\cdot(1/6)=\sqrt{5/6}/3=\sqrt{5/54}$. $m/n=5/54$, $\gcd(5,54)=1$, $m+n=59$.
Correct Answer: 59