Unit Circle

Given a point P on the unit circle, find the coordinates of P if the angle formed with the positive x-axis is θ, where θ satisfies 0 ≤ θ ≤ 2π and the coordinates satisfy the equation of the circle x^2 + y^2 = 1 Provide three different coordinate sets for

$$\theta = \frac{\pi}{6}$$

For $\theta = \frac{\pi}{6}$, the coordinates $(x,y)$ on the unit circle are given by:

$$x = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

$$y = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$

Thus, the coordinates are $$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$.

$$\theta = \frac{\pi}{4}$$

For $\theta = \frac{\pi}{4}$, the coordinates $(x,y)$ on the unit circle are given by:

$$x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$

$$y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$

Thus, the coordinates are $$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$.

$$\theta = \frac{\pi}{3}$$

For $\theta = \frac{\pi}{3}$, the coordinates $(x,y)$ on the unit circle are given by:

$$x = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

$$y = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$

Thus, the coordinates are $$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$.

Find the general solution for \( \theta \) in the equation \( \sin(\theta) = \frac{1}{2} \) using the unit circle

To find the general solution for the equation $\sin(\theta) = \frac{1}{2}$, we use the unit circle. On the unit circle, $\sin(\theta) = \frac{1}{2}$ at $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$ within one period $[0, 2\pi)$.

Therefore, the general solutions are:

$$\theta = \frac{\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$

and

$$\theta = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$

Find the coordinates of a point on the unit circle at an angle of 45 degrees

To find the coordinates of a point on the unit circle at an angle of $45^{\circ}$, we use the sine and cosine functions.

First, we convert the angle to radians:

$$45^{\circ} = \frac{45 \pi}{180} = \frac{\pi}{4}$$

Now, we find the sine and cosine of $\frac{\pi}{4}$:

$$\cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}$$

$$\sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}$$

Therefore, the coordinates of the point are:

$$\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$$

Find the Exact Values of Trigonometric Functions for Given Angles

Given the angle $\theta = \frac{5\pi}{6}$, find the exact values of $\sin \theta$, $\cos \theta$, and $\tan \theta$ using the unit circle.

First, determine the reference angle for $\theta = \frac{5\pi}{6}$. Since $\frac{5\pi}{6}$ lies in the second quadrant, its reference angle is:

$$\pi – \frac{5\pi}{6} = \frac{\pi}{6}$$

The sine, cosine, and tangent values for $\frac{\pi}{6}$ are:

$$\sin \frac{\pi}{6} = \frac{1}{2}, \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

In the second quadrant, sine is positive while cosine and tangent are negative. Therefore:

$$\sin \frac{5\pi}{6} = \frac{1}{2}, \quad \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, \quad \tan \frac{5\pi}{6} = -\frac{1}{\sqrt{3}}$$

Find the sine and cosine values for an angle of 5π/4 radians on the unit circle

To find the sine and cosine values for an angle of $\frac{5\pi}{4}$ radians, we need to locate this angle on the unit circle.

First, we recognize that $\frac{5\pi}{4}$ radians is in the third quadrant because it is more than $\pi$ radians (180 degrees) but less than $\frac{3\pi}{2}$ radians (270 degrees).

The reference angle for $\frac{5\pi}{4}$ radians is $\frac{5\pi}{4} – \pi = \frac{\pi}{4}$ radians.

In the third quadrant, the sine and cosine values are both negative. For the reference angle $\frac{\pi}{4}$, the sine and cosine values are both $\frac{\sqrt{2}}{2}$.

Therefore, the sine and cosine values for $\frac{5\pi}{4}$ radians are:

$$\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

$$\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

Determine the Quadrant of Multiple Angles on the Unit Circle

Given the angles $30^\circ$, $150^\circ$, and $240^\circ$, determine the quadrant each angle lies in on the unit circle.

1. For $30^\circ$, it is in the first quadrant because it is between $0^\circ$ and $90^\circ$.

2. For $150^\circ$, it is in the second quadrant because it is between $90^\circ$ and $180^\circ$.

3. For $240^\circ$, it is in the third quadrant because it is between $180^\circ$ and $270^\circ$.

Therefore, $30^\circ$ lies in the first quadrant, $150^\circ$ lies in the second quadrant, and $240^\circ$ lies in the third quadrant.

Determine the coordinates of a point on the unit circle given a specific angle

Given an angle \( \theta = \frac{5\pi}{6} \), find the coordinates of the corresponding point on the unit circle.

1. The angle \( \frac{5\pi}{6} \) is in the second quadrant where sine is positive and cosine is negative.

2. Using the unit circle, the coordinates for \( \theta = \frac{5\pi}{6} \) can be found using the reference angle \( \pi – \frac{5\pi}{6} = \frac{\pi}{6} \).

3. The coordinates for \( \frac{\pi}{6} \) are \( \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).

4. Since \( \frac{5\pi}{6} \) is in the second quadrant, the x-coordinate will be negative.

Hence, the coordinates are $$\left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)$$.

Find the tangent of angle π/3 using the unit circle

To find the tangent of angle $\frac{\pi}{3}$ using the unit circle, we need to find the coordinates of the point where the terminal side of the angle intersects the unit circle.

For the angle $\frac{\pi}{3}$, the coordinates on the unit circle are $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. The tangent of an angle is given by the ratio of the y-coordinate to the x-coordinate.

$$\tan\left(\frac{\pi}{3}\right) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$$

Therefore, the tangent of $\frac{\pi}{3}$ is $\sqrt{3}$.

Determine the quadrant of point P on the unit circle where P has coordinates (cos(θ), sin(θ)), given that θ = 210 degrees

First, convert the given angle to radians:

$$\theta = 210^\circ \times \frac{\pi}{180^\circ} = \frac{7\pi}{6}$$

Determine the coordinates of point P:

$$P = (\cos(210^\circ), \sin(210^\circ))$$

Since $210^\circ$ is in the third quadrant, $\cos(210^\circ)$ is negative and $\sin(210^\circ)$ is also negative. Therefore, the point P is in the third quadrant.

The answer is: The point P is in the third quadrant.

Given the point P(a, b) on the unit circle, find the exact values of sine, cosine, and tangent for the angles $\theta$ and $\phi$ where $\theta$ is the angle between the positive x-axis and the line segment OP and $\phi$ is the angle in radians correspond

Given the point $P(a, b)$ on the unit circle, we know that $a^2 + b^2 = 1$.

For angle $\theta$:

The sine and cosine values are the coordinates of point P, so:

$$\sin(\theta) = b$$

$$\cos(\theta) = a$$

To find the tangent, we use:

$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{b}{a}$$

For angle $\phi$:

Since $\phi$ represents the arc length from $(1, 0)$ to $P$, we use the unit circle property that $\phi$ forms the same angle as $\theta$ from the origin:

$$\sin(\phi) = b$$

$$\cos(\phi) = a$$

$$\tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} = \frac{b}{a}$$