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

2 Answers
May 13, 2018

Answer:

#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#

May 13, 2018

Answer:

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.