Math  /  Trigonometry

QuestionThe unit circle is shown below. Complete the following. (a) Sketch θ=30\theta=-30^{\circ} in standard position on the unit circle.
Find the lengths of the legs of its reference triangle. These are labeled aa and bb in the figure below, when an angle is sketched. Then use your reference triangle to find the coordinates of point PP. Use exact values and not decimal approximations. a=b=P=(,)\begin{array}{l} a=\square \\ b=\square \\ P=(\square, \square) \end{array}

Studdy Solution

STEP 1

1. The unit circle has a radius of 1.
2. The angle θ=30\theta = -30^\circ is measured clockwise from the positive x-axis.
3. The reference triangle is a right triangle formed by dropping a perpendicular from the terminal side of the angle to the x-axis.

STEP 2

1. Sketch the angle θ=30\theta = -30^\circ on the unit circle.
2. Identify the reference triangle and find the lengths of its legs, aa and bb.
3. Determine the coordinates of point PP using the reference triangle.

STEP 3

To sketch θ=30\theta = -30^\circ, start at the positive x-axis and rotate 30 degrees clockwise. This places the terminal side of the angle in the fourth quadrant.

STEP 4

The reference angle for 30-30^\circ is 3030^\circ. In a 30-60-90 triangle, the sides are in the ratio 1:3:21 : \sqrt{3} : 2.
Since the hypotenuse is the radius of the unit circle, which is 1, the side opposite the 3030^\circ angle (vertical leg) is 12\frac{1}{2}, and the side adjacent to the 3030^\circ angle (horizontal leg) is 32\frac{\sqrt{3}}{2}.

STEP 5

Since the angle is in the fourth quadrant, the horizontal leg aa is positive, and the vertical leg bb is negative.
Thus, a=32a = \frac{\sqrt{3}}{2} and b=12b = -\frac{1}{2}.

STEP 6

The coordinates of point PP are given by (a,b)(a, b), which are the x and y coordinates of the point on the unit circle.
Therefore, P=(32,12)P = \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right).
The lengths of the legs and the coordinates of point PP are:
a=32b=12P=(32,12)\begin{array}{l} a = \frac{\sqrt{3}}{2} \\ b = -\frac{1}{2} \\ P = \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) \end{array}

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