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I'm being asked to find the arc length of $y=sin x$ for [0, $\frac{pi}{2}$] using $M_8$.

I've determined that $y\prime^2=cos^2x$. So, using the formula for arc length, I get $\sqrt{1+cos^2x}$ as my function.

Now, they want me to evaluate this using $M_8$, so I end up with 8 midpoints from $\frac{pi}{32}$ to $\frac{15pi}{32}$ moving up in increments of $\frac{pi}{16}$.

Also, I know that $\Delta{x} = \frac{pi}{16}$. So, i end up with this estimated integral:

$\frac{pi}{16}(\sqrt{1+cos^2(\frac{pi}{32})} + \sqrt{1+cos^2(\frac{3pi}{32})} + ... \sqrt{1+cos^2(\frac{15pi}{32})}) = 1.91009889$

Please feel free to correct any errors here.

My question, however, has to do with simplifying this answer using trig. I would assume that there's a fancy way to do this, rather than just handing in the answer 1.91009889 (though the prof will accept this).

I'd like to sharpen my trig skills, so I figured this was a great chance.

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up vote 1 down vote accepted

Offhand, not much comes to mind with regard to $\sqrt{1+\cos^2x}$, but $\cos\frac{k\pi}{2^n}$ can be evaluated exactly using the half-angle identity $\cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}}$ (choose + or – based on the value of $\frac{x}{2}$) repeatedly.

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+1 Thanks for the quick response! – Stephano Mar 7 '11 at 21:40
I'm going to give you the answer. This may just be a calculator question. We do run across those in the book once in a while. Cheers! – Stephano Mar 7 '11 at 22:54
@Stephano: Thanks. Since it's asking for a midpoint sum, I would expect it to be looking for an approximation, so it's probably a calculator question. Which book are you using? – Isaac Mar 7 '11 at 22:59
Rogawski 1429210737. This question was 8.1.12. – Stephano Mar 10 '11 at 19:41

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