Question

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

Answer 1

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

Explanation:

Coordinate of P:
`x = sqrt3/2`, and `y = - 1/2` --> t is in Quadrant 4.
`tan t = y/x = (-1/2)(2/sqrt3) = - 1/sqrt3 = - sqrt3/3`
`cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 1/3) = 3/4`
`cos t = sqrt3/2` (because t is in Quadrant 4, cos t is positive)
`sin^2 t = 1 - cos^2 t = 1 - 3/4 = 1/4`
`sin t = +- 1/2`
Since t is in Quadrant 4, then, sin t is negative
`sin t = - 1/2`

Answer 2

Since `|P|^2= (sqrt{3}/2)^2 + (-1/2)^2 = 1,` we see `P` is on the unit circle so the cosine of its angle is its x coordinate, `\cos theta = sqrt{3}/2,` and the sine is its y coordinate, `sin theta = -1/2.`

Explanation:

In this problem we're only asked for `sin theta` and `cos theta,` not `theta,` so the question writer could have skipped the biggest cliche in trig, the 30/60/90 right triangle. But they just can't help themselves.

Students should recognize immediately The Two Tired Triangles of Trig. Trig mostly only uses two triangles, namely 30/60/90, whose sines and cosines in the various quadrants are `\pm 1/2` and `\pm \sqrt{3}/2` and 45/45/90, whose sines and cosines are `\pm \sqrt{2}/2 = pm 1/sqrt{2}.`

Two triangles for a whole course is really not that much to memorize. Rule of thumb: `sqrt{3}` in a problem means 30/60/90 and `\sqrt{2}` means 45/45/90.

None of that mattered for this particular problem so I'll end my rant here.