**If P(θ) is the point at the intersection of the terminal arm of angle θ and the unit circle, determine the exact coordinates of each of the following.**

a) P (\({3π \over 4}\))

I know a special triangle (specifically an isosceles right) can be used for this, but why?

The angle here is \({π \over 4}\), not \({3π \over 4}\), I don't understand why it's okay to use the isosceles right?

Thank you! :)

Guest Dec 6, 2018

#1**+1 **

\(P(\theta) = \{\cos(\theta),~\sin(\theta)\}\\ P\left(\dfrac{3\pi}{4}\right) = \left\{\cos\left(\dfrac{3\pi}{4}\right),~\sin\left(\dfrac{3\pi}{4}\right)\right\} = \left\{-\dfrac{\sqrt{2}}{2},~\dfrac{\sqrt{2}}{2}\right\}\)

Rom
Dec 6, 2018

#2**+2 **

First, realize that in Quandrant I sin and cos are positive

in Quandrant II sin is positive and cos is negative

Using the unit circle pi/4 sin = opp/hyp = 1/sqrt2 = sqrt2/2 (this is the ' y' coordinate)

cos = adj/hyp = 1/sqrt2 = sqrt 2/2 (this is the 'x' coordinate)

(in Quandrant I they are BOTH POSITIVE)

Now move to Quandrant II the magnitude of sin and cos are the same BUT cos is negative

so for 3pi/4 sin = sqrt2/2 cos = - sqrt2/2

so (x,y) on the unit circle for 3pi/4 = ( - sqrt2/2 , sqrt2/2 )

ElectricPavlov
Dec 6, 2018