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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
avatar+3213 
+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
avatar+14579 
+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

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