anonymous one year ago Session 56, first recitation - a mistake? Why is the final answer 2√2 and not 4√2?

1. 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 $-\left( - \frac{\sqrt{2}}{2}\right) -\left( - \frac{\sqrt{2}}{2}\right)+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\= \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \\= \frac{4\sqrt{2}}{2} = 2 \sqrt{2}$

2. 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.