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ad2h
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Session 56, first recitation  a mistake? Why is the final answer 2√2 and not 4√2?
 3 months ago
 3 months ago
ad2h Group Title
Session 56, first recitation  a mistake? Why is the final answer 2√2 and not 4√2?
 3 months ago
 3 months ago

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phi Group TitleBest ResponseYou've already chosen the best response.1
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 \[ \left(  \frac{\sqrt{2}}{2}\right) \left(  \frac{\sqrt{2}}{2}\right)+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\= \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \\= \frac{4\sqrt{2}}{2} = 2 \sqrt{2}\]
 3 months ago

ad2h Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 months ago
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