Question

Determine all points of intersection: y=cosx and y=sinx in the first quadrant

Answer 1

One when x =pi/4

Answer 2

http://www.wolframalpha.com/input/?i=y%3Dcosx+and+y%3Dsinx

Answer 3

Whoa that's really cool, thanks!

Answer 4

jim?

Answer 5

\[\Large y=\cos(x)\]



\[\Large \sin(x)=\cos(x)\]



\[\Large \sin^2(x)=\cos^2(x)\]



\[\Large \sin^2(x)=1-\sin^2(x)\]



\[\Large \sin^2(x)+\sin^2(x)=1\]



\[\Large 2\sin^2(x)=1\]



\[\Large \sin^2(x)=\frac{1}{2}\]



\[\Large \sin(x)=\sqrt{\frac{1}{2}}\]



\[\Large \sin(x)=\frac{\sqrt{2}}{2}\]



\[\Large x=\arcsin\left(\frac{\sqrt{2}}{2}\right)\]



\[\Large x=\frac{\pi}{4}\]



Now because we're restricted to the first quadrant, this means that there is only one solution.


So when \[\Large x=\frac{\pi}{4}\], then \[y=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\]

So when \[\Large x=\frac{\pi}{4}\], then \[y=\frac{\sqrt{2}}{2}\]


So the only solution is the ordered pair \[\left(\frac{\pi}{4},\frac{\sqrt{2}}{2}\right)\]

Answer 6

sry this editor is driving me crazy lol

Answer 7

lol i'm really grateful you're taking time to write it all out thanks so much! :)

Answer 8

It's not the ton of stuff to write, that's not so bad. It's just that the editor likes to cut off or shrink at weird times. Idk how to work it sometimes lol