Problem:
Let x and y be real numbers such that sinysinx=3 and cosycosx=21. The value of sin2ysin2x+cos2ycos2x can be expressed in the form qp, where p and q are relatively prime positive integers. Find p+q.
Solution:
Let sinysinx=A, and let cosycosx=B. Note that
sin2ysin2x=2sinycosy2sinxcosx=sinysinx⋅cosycosx=AB
Furthermore, sinx=Asiny and cosx=Bcosy. Square each of the last two equations and add the resulting equations to obtain 1=sin2x+ cos2x=A2sin2y+B2cos2y=A2(sin2y+cos2y)+(B2−A2)cos2y. Therefore cos2y=B2−A21−A2. Thus
cos2ycos2x=2cos2y−12cos2x−1=2cos2y−12B2cos2y−1=2(B2−A21−A2)−12B2(B2−A21−A2)−1=2−A2−B2B2−2A2B2+A2.
Substituting A=3 and B=21 produces sin2ysin2x=AB=23, and cos2ycos2x= −2919. Thus sin2ysin2x+cos2ycos2x=5849, and p+q=107.
The problems on this page are the property of the MAA's American Mathematics Competitions