overall accuracy: 75% record: 6 score: 6 evaluate lim h→0 (cos((4π/3)+h) - cos((4π/3)))/h √3/2 -√3/2 1/2…

Explanation:

Step1: Recall the cosine - addition formula

$\cos(A + B)=\cos A\cos B-\sin A\sin B$. So, $\cos(\frac{4\pi}{3}+h)=\cos\frac{4\pi}{3}\cos h-\sin\frac{4\pi}{3}\sin h$.

Step2: Substitute into the limit

The given limit $\lim_{h\rightarrow0}\frac{\cos(\frac{4\pi}{3}+h)-\cos(\frac{4\pi}{3})}{h}=\lim_{h\rightarrow0}\frac{\cos\frac{4\pi}{3}\cos h-\sin\frac{4\pi}{3}\sin h - \cos\frac{4\pi}{3}}{h}$. We know that $\cos\frac{4\pi}{3}=-\frac{1}{2}$ and $\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}$. So the limit becomes $\lim_{h\rightarrow0}\frac{-\frac{1}{2}\cos h+\frac{\sqrt{3}}{2}\sin h+\frac{1}{2}}{h}=\lim_{h\rightarrow0}\left(-\frac{1}{2}\frac{\cos h - 1}{h}+\frac{\sqrt{3}}{2}\frac{\sin h}{h}\right)$.

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 $-\frac{1}{2}\times0+\frac{\sqrt{3}}{2}\times1=\frac{\sqrt{3}}{2}$.

Answer:

$\frac{\sqrt{3}}{2}$