Problem:
Suppose cosx=0 and cos(x+z)=1/2. What is the smallest possible positive value of z?
Answer Choices:
A. 6π
B. 3π
C. 2π
D. 65π
E. 67π Solution:
Because cosx=0 and cos(x+z)=1/2, it follows that x=mπ/2 for some odd integer m and x+z=2nπ±π/3 for some integer n. Therefore
z=2nπ−2mπ±3π=kπ+2π±3π
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 π/2−π/3=π/6.
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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,−1).
Because cos(x+z)=1/2, the terminal side of an angle of measure x+z intersects the circle at C=(21,23) or D=(21,−23). Thus all angles of positive measure z=(x+z)−x can be measured counterclockwise from either OA or OB to either OC or OD. The smallest such angle is ∠BOD, which has measure π/6 and is attained, for example, when x=−π/2 and x+z=−π/3.
