create the given angle on the graph below by dragging the blue point to its location on the unit circle…

create the given angle on the graph below by dragging the blue point to its location on the unit circle: \\(\\theta = 1290^\\circ\\) show your work here hint: to add the square root symbol (\\(\\sqrt{\\square}\\)), type \root\

create the given angle on the graph below by dragging the blue point to its location on the unit circle: \\(\\theta = 1290^\\circ\\) show your work here hint: to add the square root symbol (\\(\\sqrt{\\square}\\)), type \root\

Answer

Explanation:

Step1: Find coterminal angle

To find the coterminal angle of (1290^\circ), we subtract multiples of (360^\circ) until we get an angle between (0^\circ) and (360^\circ). Let (n) be the number of full rotations. We solve (1290 - 360n) for (n) such that the result is in ([0, 360)).

First, divide (1290) by (360): (1290\div360 = 3.583\cdots). So we take (n = 3) (since (3\times360 = 1080)). Then (1290 - 3\times360=1290 - 1080 = 210^\circ). Wait, let's check with (n = 4): (4\times360 = 1440), which is more than (1290), so (n = 3) gives (210^\circ)? Wait, no, wait: (360\times3 = 1080), (1290 - 1080 = 210)? Wait, no, (360\times3 = 1080), (1290 - 1080 = 210)? Wait, but (360\times3 = 1080), (1290 - 1080 = 210). Wait, but let's check again: (360\times3 = 1080), (1290 - 1080 = 210). But wait, (210^\circ) is in the third quadrant. Wait, but maybe I made a mistake. Let's do it properly:

The formula for coterminal angles is (\theta - 360k), where (k) is an integer, such that (0\leq\theta - 360k<360).

So we need to find (k) such that (0\leq1290 - 360k<360).

Let's solve for (k):

(1290 - 360k\geq0\Rightarrow360k\leq1290\Rightarrow k\leq\frac{1290}{360}\approx3.583)

(1290 - 360k<360\Rightarrow - 360k<360 - 1290\Rightarrow - 360k<-930\Rightarrow k>\frac{930}{360}\approx2.583)

So (k = 3) (since (k) must be integer). Then (1290 - 3\times360 = 1290 - 1080 = 210^\circ). Wait, but (210^\circ) is (180 + 30^\circ), so in the third quadrant. But wait, let's check with (k = 4): (1290 - 4\times360 = 1290 - 1440=-150^\circ), which is negative, so not in the range. So the coterminal angle is (210^\circ). Wait, but maybe I made a mistake. Wait, (360\times3 = 1080), (1290 - 1080 = 210). Yes.

Wait, but let's check the unit circle. The angle (210^\circ) is in the third quadrant, with reference angle (30^\circ) (since (210 - 180 = 30)). So the coordinates on the unit circle for (210^\circ) are (\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right))? Wait, no: (210^\circ) is (180 + 30^\circ), so cosine is (-\cos30^\circ = -\frac{\sqrt{3}}{2}), sine is (-\sin30^\circ = -\frac{1}{2}). Wait, but maybe I made a mistake in the coterminal angle. Wait, let's recalculate:

Wait, (360\times3 = 1080), (1290 - 1080 = 210). Correct. So the angle (1290^\circ) is coterminal with (210^\circ), so we need to find the point on the unit circle corresponding to (210^\circ).

Wait, but let's check again: (360\times3 = 1080), (1290 - 1080 = 210). Yes. So the terminal side of (1290^\circ) is the same as (210^\circ), which is in the third quadrant, (30^\circ) below the negative x - axis.

Step2: Locate the point

The unit circle has coordinates ((\cos\theta, \sin\theta)). For (\theta = 210^\circ), (\cos210^\circ=\cos(180 + 30)^\circ=-\cos30^\circ=-\frac{\sqrt{3}}{2}\approx - 0.866), (\sin210^\circ=\sin(180 + 30)^\circ=-\sin30^\circ=-\frac{1}{2}=-0.5). So the point on the unit circle is (\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)), which is in the third quadrant, between the angles (180^\circ) and (270^\circ), (30^\circ) from the negative x - axis.

To drag the blue point (which is initially at ((1,0)) or ((0,1))? Wait, the blue point is at ((1,0)) initially? Wait, the graph shows the blue point at ((1,0)) (on the positive x - axis). So we need to move it to the point corresponding to (210^\circ), which is (\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)\approx(-0.866, - 0.5)).

Answer:

The coterminal angle of (1290^\circ) is (210^\circ), and the point on the unit circle is (\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)). To create the angle, drag the blue point to the location (\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)) on the unit circle (in the third quadrant, (30^\circ) below the negative x - axis).