Unit Circle

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:

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

Find the coordinates of the point on the unit circle corresponding to the angle whose cosine is -2/3

Given the cosine of the angle is $-\frac{2}{3}$,

First, find the sine of the angle using the Pythagorean identity:

$$\cos^2\theta + \sin^2\theta = 1$$

Substitute $\cos\theta = -\frac{2}{3}$:

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

$$\frac{4}{9} + \sin^2\theta = 1$$

$$\sin^2\theta = 1 – \frac{4}{9}$$

$$\sin^2\theta = \frac{9}{9} – \frac{4}{9}$$

$$\sin^2\theta = \frac{5}{9}$$

Taking the square root,

$$\sin\theta = \pm\sqrt{\frac{5}{9}}$$

$$\sin\theta = \pm\frac{\sqrt{5}}{3}$$

Thus, the coordinates are:

$$(-\frac{2}{3}, \frac{\sqrt{5}}{3})$$ or $$(-\frac{2}{3}, -\frac{\sqrt{5}}{3})$$