Unit Circle

Find the angles on the unit circle where cos(θ) = -1/2

To find the angles where $\cos(\theta) = -\frac{1}{2}$ on the unit circle, we need to consider the unit circle properties and the cosine function.

1. The cosine of an angle represents the x-coordinate on the unit circle.

2. $\cos(\theta) = -\frac{1}{2}$ corresponds to the x-coordinate -1/2.

3. The angles with $\cos(\theta) = -\frac{1}{2}$ are in the second and third quadrants because cosine is negative in these quadrants.

4. The reference angle for $\cos(\theta) = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$.

5. Therefore, the angles are:

$\theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3}$

$\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

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

What is the cosine of 30 degrees on the unit circle?

To find the cosine of $30^{\circ}$ on the unit circle, we need to recall the coordinates of the point on the unit circle corresponding to $30^{\circ}$. The coordinates are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$.

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

Therefore, the cosine of $30^{\circ}$ is $\frac{\sqrt{3}}{2}$.

Find the exact values of sin(3π/4), cos(3π/4), and tan(3π/4) using the unit circle

We are asked to find the exact values of $\sin(\frac{3\pi}{4})$, $\cos(\frac{3\pi}{4})$, and $\tan(\frac{3\pi}{4})$ using the unit circle.

First, we locate the angle $\frac{3\pi}{4}$ on the unit circle: it is in the second quadrant.

The reference angle for $\frac{3\pi}{4}$ is $\frac{\pi}{4}$ (45 degrees). In the second quadrant, the sine value is positive, and the cosine value is negative.

Thus, we have:

$$\sin(\frac{3\pi}{4}) = \sin(\pi – \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$

$$\cos(\frac{3\pi}{4}) = \cos(\pi – \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$

$$\tan(\frac{3\pi}{4}) = \frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1$$

Find the angle on the unit circle in the complex plane where the cosine value is 1/2

We start by knowing that the cosine function gives the real part of the point on the unit circle corresponding to a given angle.

We are given $\cos(\theta) = \frac{1}{2}$ and need to find the angles $\theta$ where this holds true.

On the unit circle, $\cos(\theta)$ reaches $\frac{1}{2}$ at two points: $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$.

Therefore, the angles are:

$$\theta = \frac{\pi}{3}, \frac{5\pi}{3}$$

Find the sine and cosine values for the angle θ = 30° on the unit circle

To find the sine and cosine values for $ \theta = 30° $ on the unit circle, we need to locate the point on the unit circle corresponding to $ \theta = 30° $.

1. Convert degrees to radians, $ \theta = 30° = \frac{π}{6} $ radians.

2. From the unit circle, the coordinates for $ \theta = \frac{π}{6} $ are $ ( \cos(30°), \sin(30°) ) $.

3. Hence, $ \cos(30°) = \frac{\sqrt{3}}{2} $ and $ \sin(30°) = \frac{1}{2} $.

So, the cosine value is $ \frac{\sqrt{3}}{2} $ and the sine value is $ \frac{1}{2} $.

Find the value of tan(θ) for θ in the unit circle

To find the value of $\tan(\theta)$ for $\theta$ on the unit circle, consider the point $P(x, y)$ on the circle corresponding to the angle $\theta$. The tangent of $\theta$ is given by the ratio of the y-coordinate to the x-coordinate, i.e., $\tan(\theta) = \frac{y}{x}$. For example, if $\theta = \frac{\pi}{4}$, the coordinates of the corresponding point on the unit circle are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$. Therefore,

$$\tan \left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1.$$

Determine the Quadrant of a Given Point on a Unit Circle

Given the point \((x, y)\) on a unit circle, determine the quadrant in which the point lies.

The unit circle has a radius of 1. The quadrants are defined as follows:

– Quadrant I: \((x > 0, y > 0)\)

– Quadrant II: \((x < 0, y > 0)\)

– Quadrant III: \((x < 0, y < 0)\)

– Quadrant IV: \((x > 0, y < 0)\)

Let’s solve for the point \((-\frac{1}{2}, \frac{\sqrt{3}}{2})\)

Given: \(x = -\frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\)

Since \(x < 0\) and \(y > 0\), the point lies in Quadrant II.

Find the value of tan(θ) at three specific angles on the unit circle: θ = π/4, 3π/4, and 5π/6

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

On the unit circle, at $\theta = \frac{\pi}{4}$, the coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

The tangent function is given by $\tan(\theta) = \frac{y}{x}$.

Thus,
$$ \tan(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

For $\theta = \frac{3\pi}{4}$:

On the unit circle, at $\theta = \frac{3\pi}{4}$, the coordinates are $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

Thus,
$$ \tan(\frac{3\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

For $\theta = \frac{5\pi}{6}$:

On the unit circle, at $\theta = \frac{5\pi}{6}$, the coordinates are $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.

Thus,
$$ \tan(\frac{5\pi}{6}) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$

Convert the angle 150 degrees to radians and find its coordinates on the unit circle

To convert degrees to radians, we use the conversion factor \( \frac{\pi}{180} \).

$$150^{\circ} \times \frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6}$$

The coordinates on the unit circle for \( \theta = \frac{5\pi}{6} \) are given by \((\cos(\theta), \sin(\theta))\).

$$\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$$

Thus, the coordinates are:

$$( -\frac{\sqrt{3}}{2}, \frac{1}{2} )$$

If $\theta$ is an angle in the unit circle such that $\cos(\theta) = \frac{1}{2}$ and $\sin(\theta) = \frac{\sqrt{3}}{2}$, find the value of $\theta$ in degrees and radians

To solve this problem, we need to determine the angle $\theta$ on the unit circle. Given:

$$\cos(\theta) = \frac{1}{2}$$ $$\sin(\theta) = \frac{\sqrt{3}}{2}$$

On the unit circle, these values correspond to the angle $\theta = 60^{\circ}$ or $\theta = \frac{\pi}{3}$ radians.

Therefore, the value of $\theta$ is:

$$\theta = 60^{\circ}$$ $$\theta = \frac{\pi}{3}$$