Identify what angle, θ \theta θ , satisfies the following conditions. cos θ = 3 2 \cos\theta=\frac{\sqrt3}{2} cos θ = 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> ; sin θ \sin\theta sin θ < 0 0 0

In this problem, we're asked to identify which angle theta satisfies the following conditions. And here we're told that the cosine of theta is equal to the square root of 3 over 2 and the sine of theta is less than 0. So we wanna look for an angle on our unit circle that satisfies both of these conditions. Now, let's first look at our first condition, the cosine of theta being equal to the square root of 3. So looking at my unit circle, I know that the cosine of my angle is that x value in these coordinates. So looking in quadrant 1 first, I see that the cosine is equal to the square root of 3 over 2 for 30 degrees. But then going around my unit circle and ending up over here in quadrant 4, I see that the cosine of 30 of 330 degrees is also equal to the square root of 2. Now because of that, this is where our second condition comes in. Now our second condition is that the sine of theta must be less than 0. That tells us that the sine of our angle must be negative. Now looking at that first angle that I identified, 30 degrees, this is in quadrant 1 where all of my functions are positive. So my sine can't be negative here, so 30 degrees is out of the running. Now let's make sure that our 330 degrees does satisfy this condition. Now looking here at our sign, our sign is negative one half, which is indeed a negative number less than 0. So that tells us that our angle that satisfies both of these given conditions is 330 degrees or 11 pi over 6. So theta is equal to 330 degrees. Thanks for watching and let me know if you have questions.

9. Unit Circle

Reference Angles

Precalculus

9. Unit Circle

Reference Angles

Struggling with Precalculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Identify what angle, $\theta$ , satisfies the following conditions.
$\cos\theta=\frac{\sqrt3}{2}$ ; $\sin\theta$ < $0$

30°

60°

120°

330°

Verified step by step guidance

Understand that the problem is asking for an angle θ where the cosine of θ is equal to $ \frac{\sqrt{3}}{2} $ and the sine of θ is less than 0.

Recall that the cosine function is positive in the first and fourth quadrants. The value $ \frac{\sqrt{3}}{2} $ corresponds to angles of 30° and 330° in these quadrants.

Since the sine of θ must be less than 0, we need to focus on the fourth quadrant, where sine values are negative.

Identify that in the fourth quadrant, the angle that has a cosine of $ \frac{\sqrt{3}}{2} $ and a negative sine is 330°.

Verify that 330° satisfies both conditions: $ \cos(330°) = \frac{\sqrt{3}}{2} $ and $ \sin(330°) < 0 $.

5:31m

Watch next

Master Reference Angles on the Unit Circle with a bite sized video explanation from Patrick

Start learning

Reference Angles practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice

Struggling with Precalculus?

Identify what angle, θ\thetaθ , satisfies the following conditions.cosθ=32\cos\theta=\frac{\sqrt3}{2}cosθ=23; sinθ\sin\thetasinθ < 000

Watch next

Reference Angles practice set

Identify what angle, $\theta$ , satisfies the following conditions.
$\cos\theta=\frac{\sqrt3}{2}$ ; $\sin\theta$ < $0$