Unit Circle

Convert the point on the unit circle given in Cartesian coordinates (sqrt(3)/2, 1/2) to its corresponding angle in degrees and radians, and verify the solution by converting the angle back to Cartesian coordinates

We are given the point $(\sqrt{3}/2, 1/2)$ on the unit circle. To find the corresponding angle, we use the following trigonometric relationships:

$$x = \cos(\theta)$$

$$y = \sin(\theta)$$

Thus, we have:

$$\cos(\theta) = \sqrt{3}/2$$

$$\sin(\theta) = 1/2$$

For the angle $\theta$ that satisfies these equations, we recognize that these are standard values. The angle $\theta$ is $30^{\circ}$ or $\pi/6$ radians.

To verify, we will convert $30^{\circ}$ back to Cartesian coordinates:

$$\cos(30^{\circ}) = \sqrt{3}/2, \sin(30^{\circ}) = 1/2$$

Thus, the point $(\cos(30^{\circ}), \sin(30^{\circ})) = (\sqrt{3}/2, 1/2)$ matches the given point. Therefore, $\theta = 30^{\circ}$ or $\pi/6$ radians.

Given a point on the unit circle, determine the coordinates and verify the trigonometric identities

Let’s consider a point $P(\cos\theta, \sin\theta)$ on the unit circle where $\theta = \frac{5\pi}{6}$. To find the coordinates and verify trigonometric identities:

First, we calculate the coordinates:

$$P = (\cos \frac{5\pi}{6}, \sin \frac{5\pi}{6})$$

Using the unit circle, we know:

$$\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$$

$$\sin \frac{5\pi}{6} = \frac{1}{2}$$

Thus, the coordinates are:

$$P = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$

Next, we verify the Pythagorean identity:

$$\cos^2 \theta + \sin^2 \theta = 1$$

Substituting in the values, we get:

$$\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1$$

Which confirms that the point lies on the unit circle.

What is the cosine of the angle π/3 on the unit circle?

To find the cosine of the angle \( \frac{\pi}{3} \) on the unit circle, we need to locate this angle on the circle.

The angle \( \frac{\pi}{3} \) corresponds to 60 degrees.

On the unit circle, the coordinates of the point at angle \( \frac{\pi}{3} \) are \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \).

The cosine of an angle is the x-coordinate of the corresponding point on the unit circle.

Therefore, \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \).

Find all angles θ between 0 and 2π such that cos(θ) = -1/2

To find the angles $\theta$ such that $\cos(\theta) = -\frac{1}{2}$, we start by identifying the quadrants where $\cos(\theta)$ is negative. Cosine is negative in the second and third quadrants.

First, we find the reference angle:

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

Now, we find the angles in the second and third quadrants:

Second quadrant: $$\pi – \frac{\pi}{3} = \frac{2\pi}{3}$$

Third quadrant: $$\pi + \frac{\pi}{3} = \frac{4\pi}{3}$$

Thus, the angles are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.

Find the coordinates of a point on the unit circle at an angle of 45 degrees from the positive x-axis

To find the coordinates of a point on the unit circle, we use the trigonometric functions sine and cosine.

The angle given is $45^{\circ}$.

Using the unit circle properties:

$$x = \cos 45^{\circ} = \frac{\sqrt{2}}{2}$$

$$y = \sin 45^{\circ} = \frac{\sqrt{2}}{2}$$

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

Find the value of x such that cos(x) = -1/2 and sin(x) is negative on the unit circle

To solve for $x$ such that $\cos(x) = -\frac{1}{2}$ and $\sin(x)$ is negative on the unit circle, follow these steps:

1. Identify the angles where $\cos(x) = -\frac{1}{2}$. This occurs at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ in radians.

2. Determine which of these angles has a negative sine value. The sine function is negative in the third and fourth quadrants.

3. Since $\frac{2\pi}{3}$ is in the second quadrant and $\frac{4\pi}{3}$ is in the third quadrant, we choose $x = \frac{4\pi}{3}$.

Therefore, the value of $x$ is:

$$ x = \frac{4\pi}{3} $$

Find the exact values of sine and cosine for the angle 5π/4 using the unit circle

To find the exact values of sine and cosine for the angle $\frac{5\pi}{4}$, we start by determining in which quadrant the angle lies.

The angle $\frac{5\pi}{4}$ is in the third quadrant because $\frac{5\pi}{4} > \pi$ and $\frac{5\pi}{4} < \frac{3\pi}{2}$.

In the third quadrant, both sine and cosine are negative.

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

We know that $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

Hence, for the third quadrant:

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

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

Given the point P on the unit circle at an angle of 210 degrees, find cos(210°) and sin(210°)

To find $\cos(210^{\circ})$ and $\sin(210^{\circ})$, we start by converting the angle to radians:

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

The reference angle for $\frac{7\pi}{6}$ is $30^{\circ}$ or $\frac{\pi}{6}$.

The coordinates of a point on the unit circle at an angle $\theta$ in the third quadrant are $(-\cos(\theta), -\sin(\theta))$.

Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$, we get:

$$\cos(210^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2}$$

$$\sin(210^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}$$

Therefore, the coordinates of the point are:

$$P(\cos(210^{\circ}), \sin(210^{\circ})) = \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$

Find the coordinates of the points where the unit circle intersects the x-axis

$$\text{The unit circle has the equation } x^2 + y^2 = 1.$$

$$\text{To find the intersection with the x-axis, we set } y = 0.$$

$$x^2 + 0^2 = 1$$

$$x^2 = 1$$

$$x = \pm 1.$$

$$\text{Thus, the coordinates are } (1, 0) \text{ and } (-1, 0).$$

Given a point P on the unit circle at an angle θ, find the coordinates of P, the length of the line segment from P to the origin, and the area of the sector formed by the angle θ in the unit circle

Given a point $P$ on the unit circle at an angle $\theta$, we can determine the coordinates of $P$ as follows:

$$ P(\cos(\theta), \sin(\theta)) $$

The length of the line segment from $P$ to the origin is simply the radius of the unit circle, which is 1.

To find the area of the sector formed by the angle $\theta$, we use the formula for the area of a sector, $$ A = \frac{1}{2} r^2 \theta $$ Since the radius $r$ is 1,

$$ A = \frac{1}{2} \theta $$

Therefore, the coordinates of $P$ are $(\cos(\theta), \sin(\theta))$, the length of the line segment from $P$ to the origin is 1, and the area of the sector is $\frac{1}{2} \theta$.