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

2 Answers
Aug 6, 2017

Answer:

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.

Aug 7, 2017

Answer:

#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#