Science:Math Exam Resources/Courses/MATH110/December 2010/Question 08 (a)

MATH110 December 2010

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Question 08 (a)
Let $\displaystyle f(x)=\sin x$ and $\displaystyle g(x)=\cos x$. Find $\displaystyle f\left({\frac {\pi }{3}}\right)$, $\displaystyle g\left({\frac {\pi }{3}}\right)$, $\displaystyle f'\left({\frac {\pi }{3}}\right)$, and $\displaystyle g'\left({\frac {\pi }{3}}\right)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
To evaluate the first two values, we simply plug it in and try to use our trigonometry tricks. The second two values are similar but we need to take a derivative first.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $\displaystyle f(x)=\sin x$ and $\displaystyle g(x)=\cos x$. Find $\displaystyle f\left({\frac {\pi }{3}}\right)$, $\displaystyle g\left({\frac {\pi }{3}}\right)$, $\displaystyle f'\left({\frac {\pi }{3}}\right)$, and $\displaystyle g'\left({\frac {\pi }{3}}\right)$ First, let's take derivatives $\displaystyle f'(x)=\cos x$ and $\displaystyle g'(x)=-\sin x$. Plugging in all the values we need, we see that we need to evaluate $\displaystyle f({\tfrac {\pi }{3}})=\sin({\tfrac {\pi }{3}})$ $\displaystyle g({\tfrac {\pi }{3}})=\cos({\tfrac {\pi }{3}})$ $\displaystyle f'({\tfrac {\pi }{3}})=\cos({\tfrac {\pi }{3}})$ $\displaystyle g'({\tfrac {\pi }{3}})=-\sin({\tfrac {\pi }{3}})$ So it suffices to evaluate $\displaystyle \sin({\tfrac {\pi }{3}})$ and $\displaystyle \cos({\tfrac {\pi }{3}})$ Drawing a triangle with angles 30 degrees, 60 degrees and 90 degrees (since $\displaystyle {\tfrac {\pi }{3}}$ is 60 degrees), we see that the side lengths are 1, $\displaystyle {\sqrt {3}}$ and 2. Since the shortest side is opposite the shortest angle, the middle angle is opposite the middle side and the longest angle is opposite to the longest side, we have that $\displaystyle \sin({\tfrac {\pi }{3}})={\frac {O}{H}}={\frac {\sqrt {3}}{2}}$ and $\displaystyle \cos({\tfrac {\pi }{3}})={\frac {A}{H}}={\frac {1}{2}}$. Hence, we have $\displaystyle f({\tfrac {\pi }{3}})=\sin({\tfrac {\pi }{3}})={\frac {\sqrt {3}}{2}}$ $\displaystyle g({\tfrac {\pi }{3}})=\cos({\tfrac {\pi }{3}})={\frac {1}{2}}$ $\displaystyle f'({\tfrac {\pi }{3}})=\cos({\tfrac {\pi }{3}})={\frac {1}{2}}$ $\displaystyle g'({\tfrac {\pi }{3}})=-\sin({\tfrac {\pi }{3}})=-{\frac {\sqrt {3}}{2}}$

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