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

May 13, 2018

$\sin t = - \frac{1}{2}$
$\cos t = \frac{\sqrt{3}}{2}$

#### Explanation:

Coordinate of P:
$x = \frac{\sqrt{3}}{2}$, and $y = - \frac{1}{2}$ --> t is in Quadrant 4.
$\tan t = \frac{y}{x} = \left(- \frac{1}{2}\right) \left(\frac{2}{\sqrt{3}}\right) = - \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{1}{3}} = \frac{3}{4}$
$\cos t = \frac{\sqrt{3}}{2}$ (because t is in Quadrant 4, cos t is positive)
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{3}{4} = \frac{1}{4}$
$\sin t = \pm \frac{1}{2}$
Since t is in Quadrant 4, then, sin t is negative
$\sin t = - \frac{1}{2}$

May 13, 2018

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

#### 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 $\setminus \pm \frac{1}{2}$ and $\setminus \pm \setminus \frac{\sqrt{3}}{2}$ and 45/45/90, whose sines and cosines are $\setminus \pm \setminus \frac{\sqrt{2}}{2} = \pm \frac{1}{\sqrt{2}} .$

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 $\setminus \sqrt{2}$ means 45/45/90.

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