How Can I figure out when cosine = $\frac{2}{\pi}$?

So I'm doing Mean Value theorem homework which states $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ So I am trying to find $c$ for $f(x)=\sin x$ over the interval $[0,\frac{\pi}{2}]$. So using the Mean Value theorem I got $$f'(c)=\frac{1-0}{\frac{\pi}{2}-0}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}$$ So since $f'(x)=\cos x$ $$\cos(c)=\frac{2}{\pi}$$ but I am having trouble finding the value $c$. Am I right so far, if so how do I find $c$? Thanks in advance.

• What is it that the Mean Value theorem states, exactly? Your expression $f'(c)\frac{f(b)-f(a)}{b-a}$ is not a "statement". – Omnomnomnom Jul 14 '14 at 3:33
• It would help if you wrote the problem statement, as it is written – Omnomnomnom Jul 14 '14 at 3:34
• I think that is $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ – AsdrubalBeltran Jul 14 '14 at 3:35
• @AsdrubalBeltran thanks, that's what I meant. – Kenshin Jul 14 '14 at 3:37
• The answer isn't anything pretty. Assuming they want an answer for the value of $c$, they'll be expecting you to calculate $\cos^{-1}(2/\pi)\approx 0.881$ with a calculator. – Omnomnomnom Jul 14 '14 at 3:46

Suppose that the exercise is: Find $c$, for $f(x)=\sin{x}$ with $c\in [0,\frac{π}{2}]$, that satisfied the mean value theorem. Then
With calculator: $c=\cos^{-1}(\frac{2}{\pi})$
Without calculator: how $[0,1]\subseteq[0,\frac{π}{2}]$ you can choose $[0,1]$ then
$$f'(c)=\frac{f(1)-f(0)}{1-0}=\sin(1)$$ then
$$\cos{c}=\sin(1)$$ $$\cos{c}=\cos\left(\frac{\pi}{2}-1\right)$$ $$c=\frac{\pi}{2}-1$$