Unit Circle

Determine the values of sin(θ) and cos(θ) for θ = 5π/6

Let $θ = \frac{5π}{6}$. This angle is in the second quadrant.

To find $\sin(θ)$ and $\cos(θ)$, we use the reference angle $θ’ = π – \frac{5π}{6} = \frac{π}{6}$.

The sine and cosine of $\frac{π}{6}$ are:

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

Since the angle is in the second quadrant, $\sin(θ)$ is positive and $\cos(θ)$ is negative.

Thus,

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

On the unit circle, what is the value of sin(π/6)?

To find the value of $\sin(\frac{\pi}{6})$ on the unit circle, we start by understanding that $\frac{\pi}{6}$ radians is equivalent to 30 degrees. In a unit circle, the coordinates of the angle $\frac{\pi}{6}$ are (\(\frac{\sqrt{3}}{2}, \frac{1}{2}\)).

Therefore, $\sin(\frac{\pi}{6})$ is the y-coordinate, which is:

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

Find the value of tan(θ) given that sin(θ) = 3/5 and θ is in the second quadrant Show all steps

Given that $\sin(\theta) = \frac{3}{5}$, we know the following:

Since $\sin^2(\theta) + \cos^2(\theta) = 1$

$$\left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1$$

$$\frac{9}{25} + \cos^2(\theta) = 1$$

$$\cos^2(\theta) = 1 – \frac{9}{25}$$

$$\cos^2(\theta) = \frac{16}{25}$$

Since $\theta$ is in the second quadrant, $\cos(\theta)$ is negative:

$$\cos(\theta) = -\frac{4}{5}$$

Now, $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$

$$\tan(\theta) = \frac{\frac{3}{5}}{-\frac{4}{5}}$$

$$\tan(\theta) = -\frac{3}{4}$$

So, $\tan(\theta) = -\frac{3}{4}$.

Find the exact values of sin(θ) and cos(θ) for θ = 3π/4 using the unit circle

To find the exact values of $\sin(\theta)$ and $\cos(\theta)$ for $\theta = \frac{3\pi}{4}$, we can use the unit circle.

First, note that $\theta = \frac{3\pi}{4}$ is in the second quadrant. In the unit circle, the angle $\frac{3\pi}{4}$ corresponds to $135^\circ$.

For angles in the second quadrant, the sine value is positive and the cosine value is negative. The reference angle for $\frac{3\pi}{4}$ is $\frac{\pi}{4}$ (or $45^\circ$), where $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

Therefore, $\sin(\frac{3\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{3\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$.

So, the exact values are $\sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$.

Determine the Quadrant of a Given Angle in the Unit Circle

Given an angle of 120 degrees, determine which quadrant it lies in on the unit circle.

Step 1: Recognize that the unit circle is divided into 4 quadrants:
Quadrant I: 0° to 90°
Quadrant II: 90° to 180°
Quadrant III: 180° to 270°
Quadrant IV: 270° to 360°.

Step 2: Since 120° is between 90° and 180°, it lies in Quadrant II.

Answer: Quadrant II

Identify the Quadrant of an Angle in Radians

Given an angle of $ \frac{4\pi}{3} $ radians, determine the quadrant in which the terminal side of the angle lies.

First, recall that the unit circle is divided into four quadrants:

1. Quadrant I: $0 < \theta < \frac{\pi}{2}$

2. Quadrant II: $\frac{\pi}{2} < \theta < \pi$

3. Quadrant III: $\pi < \theta < \frac{3\pi}{2}$

4. Quadrant IV: $\frac{3\pi}{2} < \theta < 2\pi$

Here, $ \frac{4\pi}{3} $ radians is greater than $ \pi $ and less than $ \frac{3\pi}{2}$. Hence, it lies in Quadrant III.

Find the Cosine of an Angle on the Unit Circle

To find the cosine of an angle, we use the unit circle. Given that the angle is $\theta = \frac{\pi}{3}$, we need to find $\cos(\frac{\pi}{3})$.

On the unit circle, the coordinates of the point corresponding to the angle $\theta$ are $(\cos(\theta), \sin(\theta))$. For $\theta = \frac{\pi}{3}$, the coordinates are $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. So,

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

Determine the tan values of specific angles on the unit circle

We need to determine the $\tan$ values for the angles $30^{\circ}$, $45^{\circ}$, and $60^{\circ}$ on the unit circle:

1. For $30^{\circ}$:

$$\tan 30^{\circ} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

2. For $45^{\circ}$:

$$\tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$$

3. For $60^{\circ}$:

$$\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$$

What are the sine, cosine, and tangent of the angle $\frac{\pi}{4}$ on the unit circle?

To find the sine, cosine, and tangent of the angle $\frac{\pi}{4}$ on the unit circle, we need to locate the angle on the circle.

The angle $\frac{\pi}{4}$ radians is equivalent to 45 degrees. In the unit circle, this corresponds to the point where both x and y coordinates are equal, as the angle bisects the first quadrant.

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

Therefore:

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

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

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

Find the value of csc(π/3) using the unit circle

To find $\csc(\frac{\pi}{3})$, we first need to recall the definition of the cosecant function:

$$\csc(\theta) = \frac{1}{\sin(\theta)}$$

Next, we locate the angle $\frac{\pi}{3}$ on the unit circle. The sine of $\frac{\pi}{3}$ is given by:

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

Now, using the definition of cosecant:

$$\csc(\frac{\pi}{3}) = \frac{1}{\sin(\frac{\pi}{3})} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$

Therefore, $\csc(\frac{\pi}{3}) = \frac{2\sqrt{3}}{3}$.