find \\( \\lim _{h \\rightarrow 0} \\frac{\\cos \\left(\\frac{\\pi}{6}+h\\right)-\\cos \\left(\\frac{\\pi}{6}…

find \\( \\lim _{h \\rightarrow 0} \\frac{\\cos \\left(\\frac{\\pi}{6}+h\\right)-\\cos \\left(\\frac{\\pi}{6}\\right)}{h} \\).\nchoose 1 answer:\na \\( -\\frac{1}{2} \\)\nb \\( \\frac{1}{2} \\)\nc 2\nd the limit doesnt exist

find \\( \\lim _{h \\rightarrow 0} \\frac{\\cos \\left(\\frac{\\pi}{6}+h\\right)-\\cos \\left(\\frac{\\pi}{6}\\right)}{h} \\).\nchoose 1 answer:\na \\( -\\frac{1}{2} \\)\nb \\( \\frac{1}{2} \\)\nc 2\nd the limit doesnt exist

Answer

Explanation:

Step1: Use the cosine addition formula

The formula for (\cos(A + B)=\cos A\cos B-\sin A\sin B). Here (A=\frac{\pi}{6}) and (B = h), so (\cos(\frac{\pi}{6}+h)=\cos\frac{\pi}{6}\cos h-\sin\frac{\pi}{6}\sin h). The original limit (\lim_{h\rightarrow0}\frac{\cos(\frac{\pi}{6}+h)-\cos\frac{\pi}{6}}{h}=\lim_{h\rightarrow0}\frac{\cos\frac{\pi}{6}\cos h-\sin\frac{\pi}{6}\sin h-\cos\frac{\pi}{6}}{h}). Since (\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}) and (\sin\frac{\pi}{6}=\frac{1}{2}), we have (\lim_{h\rightarrow0}\frac{\frac{\sqrt{3}}{2}(\cos h - 1)-\frac{1}{2}\sin h}{h}).

Step2: Use the small - angle approximations

We know that (\lim_{h\rightarrow0}\frac{\cos h - 1}{h}=0) (because (\cos h-1=-2\sin^{2}\frac{h}{2}) and (\lim_{h\rightarrow0}\frac{\sin^{2}\frac{h}{2}}{\frac{h}{2}} = 0)) and (\lim_{h\rightarrow0}\frac{\sin h}{h}=1). [ \begin{align*} \lim_{h\rightarrow0}\frac{\frac{\sqrt{3}}{2}(\cos h - 1)-\frac{1}{2}\sin h}{h}&=\frac{\sqrt{3}}{2}\lim_{h\rightarrow0}\frac{\cos h - 1}{h}-\frac{1}{2}\lim_{h\rightarrow0}\frac{\sin h}{h}\ \end{align*} ] Substitute the limits: (\frac{\sqrt{3}}{2}\times0-\frac{1}{2}\times1).

Answer:

A. (-\frac{1}{2})