Unit Circle

Determine the sine, cosine, and tangent of the angle θ = 30° on the unit circle

To determine the sine, cosine, and tangent of $\theta = 30^\circ$ on the unit circle, we first need to recall the unit circle values for this angle.

For $\theta = 30^\circ$:

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

$$\cos(30^\circ) = \frac{\sqrt{3}}{2}$$

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

If the secant of an angle θ in the unit circle is 2, find the angle θ

Given $\sec(\theta) = 2$, we know $\sec(\theta) = \frac{1}{\cos(\theta)}$.

So, $\frac{1}{\cos(\theta)} = 2$ implies $\cos(\theta) = \frac{1}{2}$.

The cosine of $\theta$ is positive, so $\theta$ must be in the first or fourth quadrant.

Therefore, $\theta = \frac{\pi}{3}$ or $\theta = -\frac{\pi}{3}$.

Determine the exact values of trigonometric functions at specific angles on the unit circle

To find the exact values of the trigonometric functions for the angle $\frac{7\pi}{6}$:

1. Find the reference angle by subtracting $\pi$: $$\frac{7\pi}{6} – \pi = \frac{7\pi}{6} – \frac{6\pi}{6} = \frac{\pi}{6}$$

2. Determine the coordinates on the unit circle for $\frac{\pi}{6}$, which are $(\cos \frac{\pi}{6}, \sin \frac{\pi}{6})$. These are $$\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$$

3. Since $\frac{7\pi}{6}$ is in the third quadrant, both cosine and sine are negative: $$\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$$

Thus, $$\cos \left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2}$$ and $$\sin \left( \frac{7\pi}{6} \right) = -\frac{1}{2}$$

Determine the values of sine and cosine at a given angle on the unit circle

Given an angle of $\theta = \frac{5\pi}{6}$, determine the values of $\sin(\theta)$ and $\cos(\theta)$ using the unit circle.

First, recognize that $\theta = \frac{5\pi}{6}$ is located in the second quadrant.

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

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

Knowing the values from the unit circle:

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

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

Since $\theta = \frac{5\pi}{6}$ lies in the second quadrant, we have:

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

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

Find the value of sin(θ), cos(θ), and tan(θ) for θ = 5π/6

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

In the second quadrant, the sine function is positive and the cosine function is negative. The reference angle for $ \theta = \frac{5\pi}{6} $ is $ \frac{\pi}{6} $.

The sine and cosine values for $ \frac{\pi}{6} $ are $ \sin(\frac{\pi}{6}) = \frac{1}{2} $ and $ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $.

Since $ \theta = \frac{5\pi}{6} $ is in the second quadrant, we have:

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

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

To find $ \tan(\frac{5\pi}{6}) $, use the identity $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $:

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

Determine the angles that satisfy the tangent function

To determine the angles $\theta$ where $\tan(\theta) = \sqrt{3}$, we first recognize that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. From the unit circle, we know:

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

Thus, $\theta = \frac{\pi}{3}$ is one solution. Since tangent has a period of $\pi$, the general solution can be written as:

$$ \theta = \frac{\pi}{3} + n\pi $$

where $n$ is any integer.

Another solution within one period (from $0$ to $2\pi$) would be:

$$ \theta = \frac{4\pi}{3} $$

Therefore, the angles that satisfy $\tan(\theta) = \sqrt{3}$ within one period are $\theta = \frac{\pi}{3}$ and $\theta = \frac{4\pi}{3}$.

Find the Quadrant of a Point on the Unit Circle

To determine the quadrant of a point $(x, y)$ on the unit circle, we need to consider the signs of $x$ and $y$. The unit circle is centered at the origin (0,0) and has a radius of 1.

For example, let’s find which quadrant the point $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$ lies in:

1. Since $\frac{\sqrt{3}}{2} > 0$, the x-coordinate is positive.

2. Since $-\frac{1}{2} < 0$, the y-coordinate is negative.

From these observations, we know that the point lies in the fourth quadrant.

So, the point $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$ is in the fourth quadrant.

Fill in the missing value on the unit circle: What is the value of sin(θ) when cos(θ) = -1/2?

Given the unit circle equation:

$$\cos^2(θ) + \sin^2(θ) = 1$$

Substitute $\cos(θ) = -\frac{1}{2}$:

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

Simplify:

$$\frac{1}{4} + \sin^2(θ) = 1$$

Subtract $\frac{1}{4}$ from both sides:

$$\sin^2(θ) = 1 – \frac{1}{4}$$

$$\sin^2(θ) = \frac{3}{4}$$

Taking the square root of both sides:

$$\sin(θ) = \pm \sqrt{\frac{3}{4}}$$

$$\sin(θ) = \pm \frac{\sqrt{3}}{2}$$

So, the value of $\sin(θ)$ when $\cos(θ) = -\frac{1}{2}$ is $\pm \frac{\sqrt{3}}{2}$.

Find the values of sin, cos, and tan for 45 degrees on the unit circle

Given $\theta = 45^\circ$, we need to find the values of $\sin(45^\circ)$, $\cos(45^\circ)$, and $\tan(45^\circ)$.

From the unit circle, we know that $\sin(45^\circ) = \frac{\sqrt{2}}{2}$ and $\cos(45^\circ) = \frac{\sqrt{2}}{2}$.

Using the identity $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$, we get:

$$\tan(45^\circ) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$$

Therefore, $\sin(45^\circ) = \frac{\sqrt{2}}{2}$, $\cos(45^\circ) = \frac{\sqrt{2}}{2}$, and $\tan(45^\circ) = 1$.

Find the sine and cosine values for a given angle on the unit circle

Suppose we need to find the sine and cosine values for the angle $ \frac{\pi}{4} $ on the unit circle.

Step 1: Recognize the angle $ \frac{\pi}{4} $ on the unit circle.

Step 2: Recall that $ \frac{\pi}{4} $ corresponds to 45 degrees.

Step 3: Know that the coordinates at 45 degrees (or $ \frac{\pi}{4} $ radians) on the unit circle are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

Step 4: Therefore, $ \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $ and $ \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $.

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

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