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Mathematics 15 Online
OpenStudy (anonymous):

dont understand this assigment

OpenStudy (anonymous):

OpenStudy (anonymous):

jeez that's a killer maybe we can do a couple of them

OpenStudy (anonymous):

You are breaking down the sin function into words. Start with the Amplitude and work your way down

OpenStudy (anonymous):

because \(\sin(\theta)\) is the second coordinate of a point on the unit circle, whose equation is \(x^2+y^2=1\) you know the range of sine is \([-1,1]\) since the lowest point on the circle is \((0,-1)\) and the highest point is \((0,1)\)

OpenStudy (anonymous):

this also makes the "amplitude" one since it is the highest the sine function can go

OpenStudy (anonymous):

because one you traverse around the unit circle one rotation, you are right back where you started from, and because we know the unit circle has circumference \(2\pi\) we see that \(\sin(\theta)=\sin(\theta +2\pi)\) making sine a periodic function with period \(2\pi\)

OpenStudy (anonymous):

first off, do you know how sine relates to the unit circle?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

ahh ok then that is where we need to start. so you know how to find the sine of any number for example \[\sin(\frac{\pi}{2})\]?

OpenStudy (anonymous):

you do it by finding the second coordinate of the point on the unit circle corresponding to the angle

OpenStudy (anonymous):

ok it is here lets take it step by step

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