We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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! :)

 Dec 6, 2018

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

 Dec 6, 2018

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 )

 Dec 6, 2018

7 Online Users