Homework

Find the values of sine, cosine, and tangent for an angle of 120 degrees using the unit circle

To find the values of $\sin$, $\cos$, and $\tan$ for an angle of 120 degrees, first convert the angle to radians:

$$120^\circ = \frac{120 \pi}{180} = \frac{2 \pi}{3}$$

Next, locate the angle on the unit circle. The angle $\frac{2 \pi}{3}$ is in the second quadrant, where the sine is positive, and the cosine and tangent are negative.

The reference angle for $120^\circ$ is $180^\circ – 120^\circ = 60^\circ$.

For $60^\circ$, we have:

$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$

$$\cos 60^\circ = \frac{1}{2}$$

Since 120 degrees is in the second quadrant:

$$\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$$

$$\cos 120^\circ = -\cos 60^\circ = -\frac{1}{2}$$

$$\tan 120^\circ = \frac{\sin 120^\circ}{\cos 120^\circ} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$$

Finding Specific Tan Values on the Unit Circle

To find the exact $\tan$ values at specific angles on the unit circle, consider the following:

1. $\theta = \frac{\pi}{4}$
At this angle, $\tan(\theta) = \tan\left(\frac{\pi}{4}\right) = 1$

2. $\theta = \frac{2\pi}{3}$
At this angle, $\tan(\theta) = \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}$

3. $\theta = \frac{7\pi}{6}$
At this angle, $\tan(\theta) = \tan\left(\frac{7\pi}{6}\right) = \frac{1}{\sqrt{3}}$

Find the Real Part of a Complex Number on the Unit Circle

Consider a complex number $z$ on the unit circle in the complex plane. The unit circle can be represented as $|z| = 1$. If $z = e^{i\theta}$, find the real part of $z$.

Solution:

Since $z = e^{i\theta}$, we can use Euler’s formula, which states:

$$e^{i\theta} = \cos \theta + i \sin \theta$$

The real part of $z$ is therefore:

$$ \cos \theta $$

Given a point on a unit circle with coordinates (x, y) and angle θ from the positive x-axis, find the coordinates of the point after rotating by 45 degrees counterclockwise

Given the initial coordinates $(x, y)$ and angle $\theta$, the coordinates after rotating by $45^\circ$ counterclockwise can be found using the rotation matrix:

$$ \begin{bmatrix} \cos(45^\circ) & -\sin(45^\circ) \\ \sin(45^\circ) & \cos(45^\circ) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$

Since $\cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2}$, the formula becomes:

$$ \begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$

Performing the matrix multiplication, we get:

$$ \begin{bmatrix} \frac{\sqrt{2}}{2}x – \frac{\sqrt{2}}{2}y \\ \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}y \end{bmatrix} $$

Thus, the new coordinates are:

$$ \left( \frac{\sqrt{2}}{2}(x – y), \frac{\sqrt{2}}{2}(x + y) \right) $$

In which quadrant does the angle lie on the unit circle?

$$Given \ an \ angle \ of \ 150^{\circ}, \ we \ need \ to \ determine \ which \ quadrant \ it \ lies \ in.$$

$$Quadrant \, I: \ 0^{\circ} \leq \theta < 90^{\circ}$$

$$Quadrant \, II: \ 90^{\circ} \leq \theta < 180^{\circ}$$

$$Quadrant \, III: \ 180^{\circ} \leq \theta < 270^{\circ}$$

$$Quadrant \, IV: \ 270^{\circ} \leq \theta < 360^{\circ}$$

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

Determine the coordinates on the unit circle for 150 degrees and the corresponding angles in radians

First, convert $150^\circ$ to radians:

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

Next, use the radian measure to find the coordinates on the unit circle. The coordinates for an angle of $\frac{5\pi}{6}$ are:

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

Thus, the coordinates for $150^\circ$ are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

Find the coordinates on the unit circle for an angle of \(\frac{\pi}{4}\)

To find the coordinates on the unit circle for an angle of $\frac{\pi}{4}$, we need to use the unit circle definition.

The unit circle has a radius of 1, and the coordinates for any angle $\theta$ can be found using $\cos(\theta)$ and $\sin(\theta)$.

For $\theta = \frac{\pi}{4}$, we have:

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

and

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

Thus, the coordinates are:

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

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:

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