Problem:
Suppose cosx=0 and cos(x+z)=1/2. What is the smallest possible positive value of z?
Answer Choices:
A. 6Οβ
B. 3Οβ
C. 2Οβ
D. 65Οβ
E. 67Οβ Solution:
Because cosx=0 and cos(x+z)=1/2, it follows that x=mΟ/2 for some odd integer m and x+z=2nΟΒ±Ο/3 for some integer n. Therefore
z=2nΟβ2mΟβΒ±3Οβ=kΟ+2ΟβΒ±3Οβ
for some integer k. The smallest value of k that yields a positive value for z is 0 , and the smallest positive value of z is Ο/2βΟ/3=Ο/6β.
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Let O denote the center of the unit circle. Because cosx=0, the terminal side of an angle of measure x, measured counterclockwise from the positive x-axis, intersects the circle at A=(0,1) or B=(0,β1).
Because cos(x+z)=1/2, the terminal side of an angle of measure x+z intersects the circle at C=(21β,23ββ) or D=(21β,β23ββ). Thus all angles of positive measure z=(x+z)βx can be measured counterclockwise from either OA or OB to either OC or OD. The smallest such angle is β BOD, which has measure Ο/6 and is attained, for example, when x=βΟ/2 and x+z=βΟ/3.
