Homework

Find the exact value of cos(5π/6) using the unit circle

To find the exact value of $\cos(\frac{5\pi}{6})$, we first determine the location of the angle on the unit circle.

The angle $\frac{5\pi}{6}$ is in the second quadrant. In the unit circle, the cosine of an angle in the second quadrant is negative.

The reference angle for $\frac{5\pi}{6}$ is $\pi – \frac{5\pi}{6}$, which simplifies to $\frac{\pi}{6}$.

The cosine of $\frac{\pi}{6}$ is $\frac{\sqrt{3}}{2}$. Therefore, $\cos(\frac{5\pi}{6}) = – \frac{\sqrt{3}}{2}$.

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

Find the coordinates of the point on the unit circle at an angle of 240 degrees

First, we convert 240 degrees to radians since the unit circle is often used with radians. The conversion factor is $\pi$ radians = 180 degrees.

Thus, $$240^\circ = \frac{240 \cdot \pi}{180} = \frac{4\pi}{3} \text{ radians}$$

Next, we find the coordinates of the point on the unit circle at an angle of $\frac{4\pi}{3}$ radians. By using the $\cos$ and $\sin$ functions:

$$x = \cos\left(\frac{4\pi}{3}\right)$$

$$y = \sin\left(\frac{4\pi}{3}\right)$$

Since $\frac{4\pi}{3}$ is in the third quadrant, where both cosine and sine are negative:

$$\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}$$

$$\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$

Therefore, the coordinates are:

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

Find the value of cos(π/4)

To find the value of $\cos(\frac{\pi}{4})$, we must understand the unit circle. The angle $\frac{\pi}{4}$, or 45 degrees, is a special angle in the unit circle.

The coordinates of the point where the terminal side of the angle $\frac{\pi}{4}$ intersects the unit circle are $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. The $x$-coordinate represents the cosine value.

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

Find the angle on the unit circle corresponding to the coordinates (-2/3, y)

To find the angle on the unit circle corresponding to the coordinates $\left(-\frac{2}{3}, y\right)$, we need to use the Pythagorean identity:

$$x^2 + y^2 = 1$$

Since $x = -\frac{2}{3}$, we plug this value into the equation:

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

$$\frac{4}{9} + y^2 = 1$$

Subtract $\frac{4}{9}$ from both sides:

$$y^2 = 1 – \frac{4}{9}$$

$$y^2 = \frac{9}{9} – \frac{4}{9}$$

$$y^2 = \frac{5}{9}$$

Take the square root of both sides:

$$y = \pm\sqrt{\frac{5}{9}}$$

$$y = \pm\frac{\sqrt{5}}{3}$$

The coordinates are $\left(-\frac{2}{3}, \pm\frac{\sqrt{5}}{3}\right)$.

Properties and Implications of the Unit Circle

$$\text{A unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. To understand its properties, consider a point } (x, y) \text{ on the unit circle. According to the equation of a circle, we have:}$$

$$x^2 + y^2 = 1$$

$$\text{For example, if } x = \frac{1}{2}, \text{ then:}$$

$$\left( \frac{1}{2} \right)^2 + y^2 = 1$$

$$\frac{1}{4} + y^2 = 1$$

$$y^2 = 1 – \frac{1}{4}$$

$$y^2 = \frac{3}{4}$$

$$y = \pm \frac{\sqrt{3}}{2}$$

Therefore, the point $( \frac{1}{2}, \pm \frac{\sqrt{3}}{2} )$ lies on the unit circle.

How to find reference angle not on unit circle

Given an angle of 210 degrees, find the reference angle.

Step 1: Determine the quadrant in which the angle lies. 210 degrees is in the third quadrant.

Step 2: Use the formula for finding the reference angle in the third quadrant: $\theta_{ref} = \theta – 180^\circ$.

Step 3: Substitute the given angle into the formula: $\theta_{ref} = 210^\circ – 180^\circ$.

Step 4: Simplify: $$\theta_{ref} = 30^\circ$$.

Given the unit circle, find the angles θ between 0 and 2π for which the secant function sec(θ) equals 2, and provide a step-by-step explanation for your answer

We start by recalling the definition of the secant function: $$\sec(\theta) = \frac{1}{\cos(\theta)}$$. Therefore, the given condition $\sec(\theta) = 2$ translates to:

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

Solving for $\cos(\theta)$, we get:

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

Now, we need to find the angles $\theta$ in the interval $[0, 2\pi)$ such that $\cos(\theta) = \frac{1}{2}$. These angles can be found using the unit circle:

$$\theta = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2k\pi \quad \text{for integers k}$$

Considering the interval $0 \leq \theta < 2\pi$, we have:

$$\theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{5\pi}{3}$$

Therefore, the angles $\theta$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

Find the coordinates of points where the angle is 2π/3 on the unit circle

To find the coordinates of the points where the angle is $$ \frac{2\pi}{3} $$ on the unit circle, we use the unit circle definition where any point can be given by $(\cos(\theta), \sin(\theta))$.

Here, $$ \theta = \frac{2\pi}{3} $$.

Therefore, the coordinates are:

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

$$ \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2} $$

Thus, the coordinates are:

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

Determine the coordinates of the point on the unit circle for an angle of 5π/6 radians Also, find the corresponding angle in degrees

To determine the coordinates of the point on the unit circle corresponding to an angle of $\frac{5\pi}{6}$ radians, we follow these steps:

1. Convert the angle into degrees:

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

2. Find the coordinates using trigonometric functions on the unit circle:

$$x = \cos(150^\circ) = \cos(180^\circ – 30^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}$$

$$y = \sin(150^\circ) = \sin(180^\circ – 30^\circ) = \sin(30^\circ) = \frac{1}{2}$$

Thus, the coordinates of the point are $$\left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)$$

The corresponding angle in degrees is $$150^\circ$$.

Finding the Coordinates on the Unit Circle

Given an angle of $\frac{5\pi}{4}$ radians, find the coordinates of the point on the unit circle.

Solution:

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

Here, $\theta = \frac{5\pi}{4}$.

First, find $\cos(\frac{5\pi}{4})$:

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

Next, find $\sin(\frac{5\pi}{4})$:

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

Therefore, the coordinates are:

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