Unit Circle

Find the cosine value for a given angle on the unit circle

Consider an angle $\theta = \frac{\pi}{3}$ on the unit circle.

We know from trigonometry that the point corresponding to $\theta = \frac{\pi}{3}$ has coordinates $(\cos(\frac{\pi}{3}), \sin(\frac{\pi}{3}))$.

Using the unit circle values, we find

$$\cos(\frac{\pi}{3}) = \frac{1}{2}$$.

Therefore, the cosine of $\frac{\pi}{3}$ is $\frac{1}{2}$.

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

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:

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

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

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

Find the coordinates on the Unit Circle

To determine the coordinates on the unit circle corresponding to an angle of $ \frac{5\pi}{6} $, we use the trigonometric functions sine and cosine.

The cosine of $ \frac{5\pi}{6} $ corresponds to the x-coordinate, and the sine of $ \frac{5\pi}{6} $ corresponds to the y-coordinate.

Calculating these values:

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

$$ \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2} $$

So, the coordinates are:

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

Find the exact value of the trigonometric functions for the angle θ = 5π/6 using the unit circle

First, locate the angle $\theta = \frac{5\pi}{6}$ on the unit circle. This angle is in the second quadrant.

In the second quadrant, sine is positive and cosine is negative.

The reference angle for $\theta = \frac{5\pi}{6}$ is $\frac{\pi}{6}$.

From the unit circle, $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.

Therefore, $\sin(\frac{5\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{5\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$.

Hence, the exact values are:

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

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

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

Find the values of tan(θ) for specific angles on the unit circle

$$For \ θ = \frac{3π}{4}, \ we \ know \ that \ tan(θ) = \frac{sin(θ)}{cos(θ)}$$

$$sin(θ) = sin(\frac{3π}{4}) = \frac{1}{\sqrt{2}}, \ cos(θ) = cos(\frac{3π}{4}) = -\frac{1}{\sqrt{2}}$$

$$Therefore, \ tan(θ) = \frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}} = -1$$