Problem:
For 135∘<x<180∘, 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. 2−22
B. 33−6
C. 32−5
D. −43
E. 1−3 Solution:
Because 135∘<x<180∘, it follows that cosx<0<sinx and ∣sinx∣<∣cosx∣. Thus tanx<0,cotx<0, and
∣tanx∣=∣cosx∣∣sinx∣<1<∣sinx∣∣cosx∣=∣cotx∣
Therefore cotx<tanx. Moreover, cotx=sinxcosx<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 QR and PS. Thus the slope of QR is equal to the slope of PS. Thus,
cotx+sinx=tanx+cosx.
Multiplying by sinxcosx=0 and rearranging gives the equivalent identity
(cosx−sinx)(cosx+sinx−sinxcosx)=0
Because cosx−sinx=0 in the required range, it follows that cosx+sinx−sinxcosx=0. Squaring and using the fact that 2sinxcosx=sin(2x) gives 1+sin(2x)=41sin2(2x). Solving this quadratic equation in the variable sin(2x) gives sin(2x)=2±22. Because −1<sin2x<1, the only solution is sin(2x)=2−22. There is indeed such a trapezoid for x=180∘+21arcsin(2−22)≈152.031∘.