evaluate lim h→0 (cos((2π/3)+h) - cos((2π/3)))/h

Answer:

C. $-\frac{\sqrt{3}}{2}$

Explanation:

Step1: Recall the cosine addition formula

$\cos(A + B)=\cos A\cos B-\sin A\sin B$. So, $\cos(\frac{2\pi}{3}+h)=\cos\frac{2\pi}{3}\cos h-\sin\frac{2\pi}{3}\sin h$.

Step2: Substitute into the limit

The original limit $\lim_{h\rightarrow0}\frac{\cos(\frac{2\pi}{3}+h)-\cos(\frac{2\pi}{3})}{h}=\lim_{h\rightarrow0}\frac{\cos\frac{2\pi}{3}\cos h-\sin\frac{2\pi}{3}\sin h - \cos\frac{2\pi}{3}}{h}$. Since $\cos\frac{2\pi}{3}=-\frac{1}{2}$ and $\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$, we have $\lim_{h\rightarrow0}\frac{-\frac{1}{2}\cos h-\frac{\sqrt{3}}{2}\sin h+\frac{1}{2}}{h}=\lim_{h\rightarrow0}\left(-\frac{1}{2}\cdot\frac{\cos h - 1}{h}-\frac{\sqrt{3}}{2}\cdot\frac{\sin h}{h}\right)$.

Step3: Use known limits

We know that $\lim_{h\rightarrow0}\frac{\cos h - 1}{h}=0$ and $\lim_{h\rightarrow0}\frac{\sin h}{h}=1$. Substituting these values, we get $-\frac{1}{2}\times0-\frac{\sqrt{3}}{2}\times1=-\frac{\sqrt{3}}{2}$.