If t= 7π/6 find the point (x,y) on the unit circle that corresponds to the real number t.

If t= 7π/6 find the point (x,y) on the unit circle that corresponds to the real number t. Calculate sin t, cos t, and tan t. Answer in the exact form

The Correct Answer and Explanation is:

To find the point (x,y)(x, y)(x,y) on the unit circle corresponding to t=7π6t = \frac{7\pi}{6}t=67π​, we will first determine the reference angle, and then use the trigonometric functions to find sin⁡t\sin tsint, cos⁡t\cos tcost, and tan⁡t\tan ttant.

Step 1: Reference Angle

The angle t=7π6t = \frac{7\pi}{6}t=67π​ is in the third quadrant of the unit circle because it is between π\piπ and 3π/23\pi/23π/2. To find the reference angle, we subtract π\piπ from ttt:Reference angle=7π6−π=7π6−6π6=π6\text{Reference angle} = \frac{7\pi}{6} – \pi = \frac{7\pi}{6} – \frac{6\pi}{6} = \frac{\pi}{6}Reference angle=67π​−π=67π​−66π​=6π​

Thus, the reference angle is π6\frac{\pi}{6}6π​.

Step 2: Coordinates on the Unit Circle

On the unit circle, the point corresponding to the angle t=7π6t = \frac{7\pi}{6}t=67π​ will have coordinates (x,y)(x, y)(x,y) where:

• x=cos⁡tx = \cos tx=cost
• y=sin⁡ty = \sin ty=sint

The coordinates for the reference angle π6\frac{\pi}{6}6π​ are:(cos⁡π6,sin⁡π6)=(32,12)\left( \cos \frac{\pi}{6}, \sin \frac{\pi}{6} \right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)(cos6π​,sin6π​)=(23​​,21​)

Since the angle is in the third quadrant, both cos⁡t\cos tcost and sin⁡t\sin tsint are negative:x=−32,y=−12x = -\frac{\sqrt{3}}{2}, \quad y = -\frac{1}{2}x=−23​​,y=−21​

Thus, the point on the unit circle corresponding to t=7π6t = \frac{7\pi}{6}t=67π​ is:(x,y)=(−32,−12)(x, y) = \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)(x,y)=(−23​​,−21​)

Step 3: Calculate Trigonometric Functions

We now calculate sin⁡t\sin tsint, cos⁡t\cos tcost, and tan⁡t\tan ttant using the values from the unit circle.

1. Sine: The sine of t=7π6t = \frac{7\pi}{6}t=67π​ is the yyy-coordinate: sin⁡t=−12\sin t = -\frac{1}{2}sint=−21​
2. Cosine: The cosine of t=7π6t = \frac{7\pi}{6}t=67π​ is the xxx-coordinate: cos⁡t=−32\cos t = -\frac{\sqrt{3}}{2}cost=−23​​
3. Tangent: The tangent is the ratio of sine to cosine: tan⁡t=sin⁡tcos⁡t=−12−32=13=33\tan t = \frac{\sin t}{\cos t} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}tant=costsint​=−23​​−21​​=3​1​=33​​

Final Answer

• The point on the unit circle corresponding to t=7π6t = \frac{7\pi}{6}t=67π​ is (−32,−12)\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)(−23​​,−21​).
• sin⁡t=−12\sin t = -\frac{1}{2}sint=−21​
• cos⁡t=−32\cos t = -\frac{\sqrt{3}}{2}cost=−23​​
• tan⁡t=33\tan t = \frac{\sqrt{3}}{3}tant=33​​

This result provides the exact values of the trigonometric functions based on the angle t=7π6t = \frac{7\pi}{6}t=67π​, calculated using the unit circle.