Homework

Calculate the sine, cosine and tangent values for angles on the unit circle

Let’s calculate the sine, cosine, and tangent values of the angle $\frac{5\pi}{6}$ on the unit circle.

First, determine the coordinates of the angle $\frac{5\pi}{6}$.

Since $\frac{5\pi}{6} = 180^\circ – 30^\circ$, it is in the second quadrant where sine is positive, and cosine is negative.

Coordinates of $30^\circ$ are $(\cos 30^\circ, \sin 30^\circ) = (\frac{\sqrt{3}}{2}, \frac{1}{2})$.

Therefore, the coordinates of $\frac{5\pi}{6}$ are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

Thus,

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

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

For tangent,

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

In the unit circle, find the coordinates of the point corresponding to an angle of 5π/6 radians Explain your steps and reasoning

To find the coordinates of the point corresponding to an angle of $\frac{5\pi}{6}$ radians on the unit circle, we need to find the cosine and sine of the angle.

First, observe that $\frac{5\pi}{6}$ radians is in the second quadrant.

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

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

Thus, the coordinates are $(-\cos \frac{\pi}{6}, \sin \frac{\pi}{6})$.

$$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$ and $$\sin \frac{\pi}{6} = \frac{1}{2}$$.

Therefore, the coordinates are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

Find the value of csc(π/4) using the unit circle

$$ \text{Step 1: Determine the sine of } \frac{\pi}{4} $$

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

$$ \text{Step 2: Use the definition of cosecant: } csc(\theta) = \frac{1}{\sin(\theta)} $$

$$ csc(\frac{\pi}{4}) = \frac{1}{\sin(\frac{\pi}{4})} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} $$

Find the general solution for cotangent of an angle on the unit circle

To find the general solution for $\cot(\theta)$ on the unit circle, first recall the definition of cotangent, which is $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$.

We are looking for the values of $\theta$ where $\cot(\theta) = k$ for some constant $k$.

Since $\cot(\theta)$ is periodic with period $\pi$, the general solution for $\theta$ can be written as:

$$\theta = \arccot(k) + n\pi$$

where $n$ is any integer. This is because the cotangent function repeats every $\pi$ radians. Thus, the full general solution for $\cot(\theta) = k$ can be written as:

$$\theta = \arccot(k) + n\pi \quad \text{for} \; n \in \mathbb{Z}$$

What is the value of sin(π/4) using the unit circle?

To find the value of $ \sin(\frac{\pi}{4}) $ using the unit circle, we need to identify the coordinates of the point on the unit circle corresponding to $ \frac{\pi}{4} $ radians.

The angle $ \frac{\pi}{4} $ radians is equivalent to 45 degrees. On the unit circle, the coordinates of the point that corresponds to 45 degrees are $ (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $.

Since $ \sin(\theta) $ is equal to the y-coordinate of the point on the unit circle, we have

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

Find the Coordinates of a Point on the Unit Circle

Given a unit circle, find the coordinates of the point where the terminal side of an angle $\theta = \frac{5\pi}{4}$ intersects the circle.

Solution:

To find the coordinates of the point where the terminal side of an angle $\theta = \frac{5\pi}{4}$ intersects the unit circle, we use the following formulas:
$x = \cos(\theta)$
$y = \sin(\theta)$

For $\theta = \frac{5\pi}{4}$:
$x = \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$y = \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

Therefore, the coordinates of the point are $\left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$.

Find the coordinates of a point on the negative unit circle

To find the coordinates of a point on the negative unit circle, we need to remember that the equation for a unit circle is $x^2 + y^2 = 1$. For a point on the negative unit circle, both x and y values will be negative.

Let’s take an example where $x = -\frac{1}{2}$. So,

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

Substituting $x = -\frac{1}{2}$ into the equation, we get:

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

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

$$ y^2 = 1 – \frac{1}{4} $$

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

$$ y = -\sqrt{\frac{3}{4}} $$

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

Therefore, the coordinates of the point are: $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$.

Determine the coordinates of the point on the unit circle corresponding to an angle of 5π/4 radians

To find the coordinates of the point on the unit circle corresponding to an angle of $$\frac{5\pi}{4}$$ radians, we can use the unit circle properties.

The angle $$\frac{5\pi}{4}$$ is located in the third quadrant of the unit circle. The reference angle for $$\frac{5\pi}{4}$$ is $$\pi – \frac{5\pi}{4} = \frac{\pi}{4}$$, which corresponds to a 45-degree angle.

For a point in the third quadrant, both the sine (y-coordinate) and cosine (x-coordinate) values will be negative. The coordinates of a 45-degree angle on the unit circle are (sqrt(2)/2, sqrt(2)/2). Therefore, the coordinates for the angle $$\frac{5\pi}{4}$$ will be:

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

Determine the sine and cosine of an angle in a unit circle

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

In a unit circle, the coordinates for $\frac{\pi}{4}$ are $(\cos(\frac{\pi}{4}), \sin(\frac{\pi}{4}))$.

Using the known values:

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

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

Therefore, the coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

What is the y-coordinate of the point on the unit circle at an angle of π/3?

To find the y-coordinate of the point on the unit circle at an angle of $\frac{\pi}{3}$, we use the sine function.

The sine of $\frac{\pi}{3}$ is $\frac{\sqrt{3}}{2}$.

Therefore, the y-coordinate is $$\frac{\sqrt{3}}{2}$$.