Question

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

Answer 1

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.

Answer 2

`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`