anonymous
  • anonymous
Session 56, first recitation - a mistake? Why is the final answer 2√2 and not 4√2?
OCW Scholar - Single Variable Calculus
katieb
  • katieb
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phi
  • phi
I assume you get to \[ - \cos(x) - \sin(x) \] evaluated from pi/4 to 5pi/4 we get the expression \[ - \cos\left(\frac{5 \pi}{4}\right)- \sin \left(\frac{5 \pi}{4}\right) - \left( -\cos\left(\frac{ \pi}{4}\right)- \sin \left(\frac{ \pi}{4}\right)\right) \\= - \cos\left(\frac{5 \pi}{4}\right)- \sin \left(\frac{5 \pi}{4}\right) +\cos\left(\frac{ \pi}{4}\right)+ \sin \left(\frac{ \pi}{4}\right) \] use the values: \[ \cos\left(\frac{5 \pi}{4}\right)= \sin\left(\frac{5 \pi}{4}\right)= - \frac{\sqrt{2}}{2} \\ \cos\left(\frac{\pi}{4}\right)= \sin\left(\frac{\pi}{4}\right)= \frac{\sqrt{2}}{2} \] to get \[ -\left( - \frac{\sqrt{2}}{2}\right) -\left( - \frac{\sqrt{2}}{2}\right)+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\= \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \\= \frac{4\sqrt{2}}{2} = 2 \sqrt{2}\]
anonymous
  • anonymous
Oh, sorry, I was confusing the trigonometric values to be square root of two instead of square root of two over two. Silly lack of attention.

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