Problem 368: (2006 AMC 12A, 15)
Suppose $\cos x=0$ and $\cos (x+z)=1/2$. What is the smallest possible positive value of $z$? $\mathrm{(A) \ } \frac{\pi}{6}\qquad$ $ \mathrm{(B) \ } \frac{\pi}{3}\qquad$ $ \mathrm{(C) \ } \frac{\pi}{2}\qquad$ $ \mathrm{(D) \ } \frac{5\pi}{6}\qquad$ $ \mathrm{(E) \ } \frac{7\pi}{6}$
Solution
- $\cos x=0$ means $x=\frac\pi2+k\pi$. Try $x=-\frac\pi2$: $\cos\left(-\frac\pi2+z\right)=\sin z=\frac12\Rightarrow z=\frac\pi6$ (smallest positive).
- Trying $x=\frac\pi2$ instead gives $-\sin z=\frac12\Rightarrow z=\frac{7\pi}6$ at smallest, which is larger.
- The overall smallest positive $z$ is (A) $\frac\pi6$.
Suppose $\sin x=0$ and $\sin(x+z)=\frac{\sqrt3}2$. What is the smallest possible positive value of $z$?
