Question

find \\( \lim _{h \
ightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\
ight)-\cos \left(\frac{\pi}{6}\
ight)}{h} \\).
choose 1 answer:
a \\( -\frac{1}{2} \\)
b \\( \frac{1}{2} \\)
c 2
d the limit doesnt exist

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 \(\lim_{h
ightarrow0}\frac{\cos(\frac{\pi}{6}+h)-\cos\frac{\pi}{6}}{h}=\lim_{h
ightarrow0}\frac{\cos\frac{\pi}{6}\cos h-\sin\frac{\pi}{6}\sin h-\cos\frac{\pi}{6}}{h}\).
Since \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\) and \(\sin\frac{\pi}{6}=\frac{1}{2}\), we have \(\lim_{h
ightarrow0}\frac{\frac{\sqrt{3}}{2}(\cos h - 1)-\frac{1}{2}\sin h}{h}\).

Step2: Use the small - angle approximations

We know that \(\lim_{h
ightarrow0}\frac{\cos h - 1}{h}=0\) (because \(\cos h-1=-2\sin^{2}\frac{h}{2}\) and \(\lim_{h
ightarrow0}\frac{\sin^{2}\frac{h}{2}}{\frac{h}{2}} = 0\)) and \(\lim_{h
ightarrow0}\frac{\sin h}{h}=1\).

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Substitute the limits: \(\frac{\sqrt{3}}{2}\times0-\frac{1}{2}\times1\).

Answer:

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