Problem:
Let x be a real number such that sin10x+cos10x=3611. Then sin12x+cos12x=nm where m and n are relatively prime positive integers. Find m+n.
Solution:
Let c=sin2x⋅cos2x, and let S(n)=sin2nx+cos2nx. Then for n≥1
S(n)=(sin2nx+cos2nx)⋅(sin2x+cos2x)=sin2n+2x+cos2n+2x+sin2x⋅cos2x(sin2n−2x+cos2n−2x)=S(n+1)+cS(n−1)
Because S(0)=2 and S(1)=1, it follows that S(2)=1−2c,S(3)=1−3c, S(4)=2c2−4c+1, and 3611=S(5)=5c2−5c+1. Hence c=61 or 65, and because 4c=sin22x, the only possible value of c is 61. Therefore
S(6)=S(5)−cS(4)=3611−61(2(61)2−4(61)+1)=5413
The requested sum is 13+54=67.
The problems on this page are the property of the MAA's American Mathematics Competitions