Trig Functions with the Unit Circle Exercise 5

Answer: \(\theta = \dfrac{\pi}{3}, \dfrac{2\pi}{3}\)

Solution Method: If we recall the Unit Circle, we can solve this problem without a calculator. 



First, we're going to look at the trig function and its sign. Each trig function is only positive in two quadrants on the unit circle, and negative in the other two. To help us remember which quadrants each trig function is positive in, we use the acronym ASTC, Always-Study-Trig-Carefully. You may have heard it before as All-Students-Take-Calculus or Add-Sugar-To-Coffee, use whatever works for you. In the first quadrant, all three (sin, cosine, and tangent) are positive. In the second quadrant, sine is positive and cosine and tangent are negative, in the third tangent is positive (because negative divided by negative is positive), and in the fourth quadrant cosine is positive. 
 
In this problem, we have a positive sine. So that narrows us down to angles in Q1 or Q2. 

Next, since \(\sin \theta =y,\) we need to recall if multiples of \(\frac{\pi}{6},\frac{\pi}{4},\) or \(\frac{\pi}{3}\) have y-coordinates of \(\pm\frac{\sqrt{3}}{2}\). The multiples of \(\frac{\pi}{3}\) have y-coordinates of \(\pm\frac{\sqrt{3}}{2}\).
 
So now we combine the facts that \(\theta\) is in Q1 or Q2 and a multiple of \(
    \frac{\pi}{3}\). The multiple of \(
    \frac{\pi}{3}\) in Q1 is \(
    \frac{\pi}{3}\), and the multiple in Q2 is \(
    \frac{2\pi}{3}\). 

So the \(\theta\)s in \([0,2\pi]\) satisfying \(\sin{\theta}=\frac{\sqrt{3}}{2}\), are \(\theta = 
    \frac{\pi}{3}\) and \(\theta = \frac{2\pi}{3}\).