Unit Circle

In which quadrant does the angle 150° lie in the unit circle?

To determine the quadrant where the angle $150^{\circ}$ lies, we divide the unit circle into four quadrants:

1. Quadrant I: $0^{\circ}$ to $90^{\circ}$

2. Quadrant II: $90^{\circ}$ to $180^{\circ}$

3. Quadrant III: $180^{\circ}$ to $270^{\circ}$

4. Quadrant IV: $270^{\circ}$ to $360^{\circ}$

Since $150^{\circ}$ is between $90^{\circ}$ and $180^{\circ}$, it lies in Quadrant II.

Find the coordinates of the point on the unit circle at 45 degrees

To find the coordinates of a point on the unit circle at a given angle, we use the angle to find the cosine and sine values, which correspond to the x and y coordinates, respectively.

For an angle of $45^{\circ}$, we have:

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

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

Thus, the coordinates of the point are:

$$(\cos(45^{\circ}), \sin(45^{\circ})) = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$

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

To find the sine of 30 degrees on the unit circle, we need to identify the coordinates of the point on the unit circle corresponding to this angle.

In the unit circle, each point is represented as $(\cos \theta, \sin \theta)$. For an angle of $30^{\circ}$, the coordinates are given by:

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

Therefore, the sine of $30^{\circ}$ is:

$$\sin 30^{\circ} = \frac{1}{2}$$

What are the sine and cosine values of 45 degrees on the unit circle?

First, let’s convert 45 degrees into radians using the conversion factor $\pi / 180$.

$$\text{Radians} = 45 \times \frac{\pi}{180} = \frac{\pi}{4}$$

On the unit circle, the coordinates corresponding to an angle of $\frac{\pi}{4}$ radians are given by $\left(\cos \frac{\pi}{4}, \sin \frac{\pi}{4}\right)$.

We know that:

$$\cos \frac{\pi}{4} = \cos 45^{\circ} = \frac{\sqrt{2}}{2}$$

$$\sin \frac{\pi}{4} = \sin 45^{\circ} = \frac{\sqrt{2}}{2}$$

Therefore, the sine and cosine values for 45 degrees on the unit circle are $\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{2}}{2}$ respectively.

Find the value of sin(θ) for θ = 7π/6 using the unit circle

To find $\sin(\theta)$ for $\theta = \frac{7\pi}{6}$, we need to locate the angle on the unit circle.

First, note that $\frac{7\pi}{6}$ is in the third quadrant where sine is negative.

$\frac{7\pi}{6}$ is $30^\circ$ past $\pi$ (180 degrees).

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

In the third quadrant, the sine of $\frac{\pi}{6}$ is $-\frac{1}{2}$.

Thus, $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$.

Determine Specific Coordinates on the Unit Circle

Given a point on the unit circle defined by the angle $\theta$ in radians, find the coordinates of the point where $\theta = \frac{5\pi}{6}$.

First, recall that any point on the unit circle is defined by $(\cos \theta, \sin \theta)$. Here $\theta = \frac{5\pi}{6}$.

So, we need to find $\cos \left(\frac{5\pi}{6}\right)$ and $\sin \left(\frac{5\pi}{6}\right)$:

• For $\cos \left(\frac{5\pi}{6}\right)$: Since $\frac{5\pi}{6}$ is in the second quadrant, $\cos \left(\frac{5\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
• For $\sin \left(\frac{5\pi}{6}\right)$: $\sin$ is positive in the second quadrant, so $\sin \left(\frac{5\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Thus, the coordinates for $\theta = \frac{5\pi}{6}$ are $$\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).$$

Find the sine, cosine, and tangent values for the angle $\theta = \frac{\pi}{4}$ on the unit circle

To find the sine, cosine, and tangent values for $\theta = \frac{\pi}{4}$, we use the unit circle properties.

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

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

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

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

Find the tangent line to the unit circle at the point (1, 0)

The equation of the unit circle is $x^2 + y^2 = 1$.

The point (1,0) is on the unit circle.

The slope of the tangent line at any point (a,b) on the circle can be found by implicit differentiation:

$$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = – \frac{x}{y}$$

At the point (1,0), the slope is undefined (vertical line).

Therefore, the equation of the tangent line is $$ x = 1 $$

Identify the quadrant of an angle

Given an angle of $135°$, determine the quadrant in which this angle lies on the unit circle.

The unit circle is divided into four quadrants:

Quadrant I: $0°$ to $90°$

Quadrant II: $90°$ to $180°$

Quadrant III: $180°$ to $270°$

Quadrant IV: $270°$ to $360°$

Since $135°$ is greater than $90°$ and less than $180°$,

$$135°$$

lies in Quadrant II.

Find the value of tan(θ) on the unit circle where θ is a special angle

To find the value of $\tan(\theta)$ on the unit circle, we need to use the relationship $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.

Let’s consider $\theta = \frac{\pi}{4}$. On the unit circle, $\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

Thus,

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

So, $\tan\left(\frac{\pi}{4}\right) = 1$.