Homework

Find the coordinates of the points where the unit circle intersects the x-axis

$$\text{The unit circle has the equation } x^2 + y^2 = 1.$$

$$\text{To find the intersection with the x-axis, we set } y = 0.$$

$$x^2 + 0^2 = 1$$

$$x^2 = 1$$

$$x = \pm 1.$$

$$\text{Thus, the coordinates are } (1, 0) \text{ and } (-1, 0).$$

Given a point P on the unit circle at an angle θ, find the coordinates of P, the length of the line segment from P to the origin, and the area of the sector formed by the angle θ in the unit circle

Given a point $P$ on the unit circle at an angle $\theta$, we can determine the coordinates of $P$ as follows:

$$ P(\cos(\theta), \sin(\theta)) $$

The length of the line segment from $P$ to the origin is simply the radius of the unit circle, which is 1.

To find the area of the sector formed by the angle $\theta$, we use the formula for the area of a sector, $$ A = \frac{1}{2} r^2 \theta $$ Since the radius $r$ is 1,

$$ A = \frac{1}{2} \theta $$

Therefore, the coordinates of $P$ are $(\cos(\theta), \sin(\theta))$, the length of the line segment from $P$ to the origin is 1, and the area of the sector is $\frac{1}{2} \theta$.

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:

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