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! :)
+6248 \(P(\theta) = \{\cos(\theta),~\sin(\theta)\}\\ P\left(\dfrac{3\pi}{4}\right) = \left\{\cos\left(\dfrac{3\pi}{4}\right),~\sin\left(\dfrac{3\pi}{4}\right)\right\} = \left\{-\dfrac{\sqrt{2}}{2},~\dfrac{\sqrt{2}}{2}\right\}\)
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+36916 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 )
