Problem:
For 135β<x<180β, points P=(cosx,cos2x),Q=(cotx,cot2x),R=(sinx,sin2x), and S=(tanx,tan2x) are the vertices of a trapezoid. What is sin(2x)?
Answer Choices:
A. 2β22β
B. 33ββ6
C. 32ββ5
D. β43β
E. 1β3β Solution:
Because 135β<x<180β, it follows that cosx<0<sinx and β£sinxβ£<β£cosxβ£. Thus tanx<0,cotx<0, and
Therefore cotx<tanx. Moreover, cotx=sinxcosxβ<cosx. Thus the four vertices P,Q,R, and S are located on the parabola y=x2 and P and S are in between Q and R. If AB and CD are chords on the parabola y=x2 such that the x-coordinates of A and B are less than the x-coordinates of C and D, then the slope of AB is less than the slope of CD. It follows that the two parallel sides of the trapezoid must be QRβ and PS. Thus the slope of QRβ is equal to the slope of PS. Thus,
cotx+sinx=tanx+cosx.
Multiplying by sinxcosxξ =0 and rearranging gives the equivalent identity
(cosxβsinx)(cosx+sinxβsinxcosx)=0
Because cosxβsinxξ =0 in the required range, it follows that cosx+sinxβsinxcosx=0. Squaring and using the fact that 2sinxcosx=sin(2x) gives 1+sin(2x)=41βsin2(2x). Solving this quadratic equation in the variable sin(2x) gives sin(2x)=2Β±22β. Because β1<sin2x<1, the only solution is sin(2x)=2β22ββ. There is indeed such a trapezoid for x=180β+21βarcsin(2β22β)β152.031β.