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

MATH110 December 2010

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[hide] 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 $\displaystyle g'\left({\frac {\pi }{3}}\right)$ .

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[show] 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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[show] Solution
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 $\displaystyle f({\tfrac {\pi }{3}})=\sin({\tfrac {\pi }{3}})={\frac {\sqrt {3}}{2}}$ $\displaystyle g({\tfrac {\pi }{3}})=\cos({\tfrac {\pi }{3}})={\frac {1}{2}}$ $\displaystyle f'({\tfrac {\pi }{3}})=\cos({\tfrac {\pi }{3}})={\frac {1}{2}}$ $\displaystyle g'({\tfrac {\pi }{3}})=-\sin({\tfrac {\pi }{3}})=-{\frac {\sqrt {3}}{2}}$