Given the point #P(sqrt2/2, sqrt2/2)#, how do you find #sintheta# and #costheta#?

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Given the point #P(sqrt2/2, sqrt2/2)#, how do you find #sintheta# and #costheta#?

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Answer 1 · 2017-01-16T19:28:20

Cal P (sqrt2/2, sqrt2/2) the terminal point of the arc t.
The radius of the circle will be:
#OP = sqrt(2/4 + 2/4) = sqrt(4/4) = 2/2 = 1#
That is the radius of the unit circle. There for:
#sin t = (sqrt3/2)#
#cos t = (sqrt3)/2#
tan t = 1
cot = 1
#sec t = 2/(sqrt3) = (2sqrt3)/3#
#csc t = 2/(sqrt3) = (2sqrt3)/3#

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Given the point #P(sqrt2/2, sqrt2/2)#, how do you find #sintheta# and #costheta#?

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