Given the point #P(-1/2, sqrt3/2)#, how do you find #sintheta# and #costheta#?

2 Answers
Zach S.
Aug 6, 2017

There is a pattern...

Explanation:

Assuming you're talking about the unit circle:

#sintheta# is the y coordinate.

#costheta# is the x coordinate.

Hope this helps! The pattern is also similar for secant and cosecant if you need help with that too! Feel free to comment back if you need more help.

Nghi N
Aug 7, 2017

#sin t = 1/2#
#cos t = - sqrt3/2#

Explanation:

#tan t = x/y = (-1/2)/(sqrt3/2) = -1/sqrt3 = - sqrt3/3#
#cos^2 t = 1/(1 + tan^2 t) = 1/( 1 + 1/3) = 3/4#
#cos t = +- sqrt3/2#
Since t is in Quadrant 3, cos t is negative -->
#cos t = - sqrt3/2#
#sin t = cos t.tan t = (- sqrt3/2)(- sqrt3/3) = 1/2#

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