Homework

Find the angle that corresponds to a given point on the unit circle

Let’s consider the point (\frac{\sqrt{3}}{2}, \, \frac{1}{2}) on the unit circle. This point lies in the first quadrant and has coordinates (cos(\theta), sin(\theta)). We need to find the angle \theta that corresponds to this point.

Using the coordinates, we know that

$$ \cos(\theta) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(\theta) = \frac{1}{2} $$

The angle \theta that satisfies both these conditions is

$$ \theta = \frac{\pi}{6} $$

Therefore, the angle corresponding to the point (\frac{\sqrt{3}}{2}, \frac{1}{2}) is \frac{\pi}{6} radians.

Find the values of θ where cot(θ) = 1 on the unit circle for 0 ≤ θ < 2π

To solve for the values of $\theta$ where $\cot(\theta) = 1$ on the unit circle for $0 \leq \theta < 2\pi$, we start by recalling that $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$. Hence, $\cot(\theta) = 1$ implies $\frac{\cos(\theta)}{\sin(\theta)} = 1$, or $\cos(\theta) = \sin(\theta)$.

On the unit circle, the equation $\cos(\theta) = \sin(\theta)$ holds when $\theta = \frac{\pi}{4} + k\pi$ for integer $k$. We need the values of $\theta$ in the interval $0 \leq \theta < 2\pi$. Thus, the possible values of $\theta$ are $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.

Therefore, the values of $\theta$ where $\cot(\theta) = 1$ on the unit circle for $0 \leq \theta < 2\pi$ are:

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

Find the value of sin(θ) and cos(θ) for θ = 45° on the unit circle

To find $\sin(45^\circ)$ and $\cos(45^\circ)$, we can use the unit circle properties.

On the unit circle, the angle $45^\circ$ (or $\frac{\pi}{4}$ radians) corresponds to the point $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

Therefore:

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

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

Given a point on the unit circle, find its cosine and sine values

Given a point \((\cos\theta, \sin\theta)\) on the unit circle, determine the coordinates when \(\theta = \frac{\pi}{4}\).

The unit circle has a radius of 1. At \(\theta = \frac{\pi}{4}\), both x and y coordinates are equal:

$$\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

Therefore, the coordinates are:

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

Find the Sine, Cosine, and Tangent Values

Consider a point on the unit circle at an angle of $\theta = 45°$.

We know that:

$$\sin(45°) = \frac{\sqrt{2}}{2}$$

$$\cos(45°) = \frac{\sqrt{2}}{2}$$

$$\tan(45°) = \frac{\sin(45°)}{\cos(45°)} = 1$$

Thus, the sine, cosine, and tangent values of 45° are $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{2}}{2}$, and 1 respectively.

Find the value of angle θ in degrees such that cos(θ) = sin(2θ) and θ lies in the interval [0, 360)

Given the equation:

$$\cos(\theta) = \sin(2\theta)$$

We can use the double-angle identity for sine:

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$

The equation becomes:

$$\cos(\theta) = 2\sin(\theta)\cos(\theta)$$

Dividing both sides by $\cos(\theta)$ (assuming $\cos(\theta) \neq 0$):

$$1 = 2\sin(\theta)$$

Solving for $\sin(\theta)$:

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

The values of $\theta$ in the interval [0, 360) where $\sin(\theta) = \frac{1}{2}$ are $\theta = 30^\circ$ and $\theta = 150^\circ$.

However, we also need to consider the case where $\cos(\theta) = 0$:

$\cos(\theta) = 0$ for $\theta = 90^\circ$ and $\theta = 270^\circ$.

Therefore, the angles that satisfy the equation are: $30^\circ$, $90^\circ$, $150^\circ$, and $270^\circ$.

Find the coordinates of the point on the unit circle that corresponds to an angle of 7π/6 radians

To find the coordinates of the point on the unit circle that corresponds to an angle of $\frac{7\pi}{6}$ radians, we can use the unit circle definitions.

The angle $\frac{7\pi}{6}$ radians is in the third quadrant where both x and y coordinates are negative.

First, we need to find the reference angle, which is $\pi – \frac{7\pi}{6} = \frac{\pi}{6}$ radians.

The coordinates corresponding to the reference angle $\frac{\pi}{6}$ are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$.

Since $\frac{7\pi}{6}$ is in the third quadrant, both coordinates will be negative. Thus, the coordinates at $\frac{7\pi}{6}$ will be:

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

Find the coordinates of the point on the unit circle at a given angle

To find the coordinates of the point on the unit circle at an angle $\theta$:

1. Use the parametric equations for the unit circle:

$$x = \cos(\theta)$$

$$y = \sin(\theta)$$

2. Substitute the given angle $\theta = \frac{2\pi}{3}$ into the equations:

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

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

Thus, the coordinates of the point are:

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

Find the point on the unit circle where the angle is π/3 and show all steps to verify the trigonometric coordinates

To find the point on the unit circle where the angle is $\frac{\pi}{3}$, we start by noting that the unit circle has a radius of 1. The coordinates of any point on the unit circle can be found using the trigonometric functions cosine (cos) and sine (sin).

We know that for an angle $\theta$:

$$ x = \cos(\theta) $$

$$ y = \sin(\theta) $$

For $\theta = \frac{\pi}{3}$:

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

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

Therefore, the coordinates of the point on the unit circle where the angle is $\frac{\pi}{3}$ are:

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

Find the coordinates of the point that results from a 225-degree rotation counterclockwise around the origin on the unit circle

To find the coordinates of a point on the unit circle after a $225^\circ$ rotation counterclockwise, we can use the trigonometric functions cosine and sine:

The general formula for finding the coordinates $(x, y)$ on the unit circle is:

$$x = \cos(\theta)$$

$$y = \sin(\theta)$$

For $\theta = 225^\circ$:

$$x = \cos(225^\circ)$$

$$y = \sin(225^\circ)$$

Since $225^\circ = 180^\circ + 45^\circ$, we can use reference angles:

$$\cos(225^\circ) = \cos(180^\circ + 45^\circ) = -\cos(45^\circ) = -\frac{\sqrt{2}}{2}$$

$$\sin(225^\circ) = \sin(180^\circ + 45^\circ) = -\sin(45^\circ) = -\frac{\sqrt{2}}{2}$$

Therefore, the coordinates are:

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