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How would I determine the exact coordinates?

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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! :)

Dec 6, 2018

#1
+5766
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$$P(\theta) = \{\cos(\theta),~\sin(\theta)\}\\ P\left(\dfrac{3\pi}{4}\right) = \left\{\cos\left(\dfrac{3\pi}{4}\right),~\sin\left(\dfrac{3\pi}{4}\right)\right\} = \left\{-\dfrac{\sqrt{2}}{2},~\dfrac{\sqrt{2}}{2}\right\}$$

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Dec 6, 2018
#2
+18839
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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