Unit Circle

Find all circle equations that lie on the unit circle

The unit circle is defined by the equation:

$$x^2 + y^2 = 1$$

To find all circles that lie on the unit circle, we consider the general equation of a circle:

$$ (x – a)^2 + (y – b)^2 = r^2 $$

For the circle to lie on the unit circle, the radius of this circle must be zero, as any larger radius would extend beyond the unit circle. Therefore:

$$ r = 0 $$

Thus, the equation simplifies to a point:

$$ (x – a)^2 + (y – b)^2 = 0 $$

Expanding this gives:

$$ x = a, y = b $$

But since it must lie on the unit circle:

$$ a^2 + b^2 = 1 $$

So, all such points (a, b) lie on the unit circle.

Therefore, the equations of all circles on the unit circle are:

$$ (x – a)^2 + (y – b)^2 = 0 $$ where $$ a^2 + b^2 = 1 $$

Find the coordinates of the point on the unit circle at an angle of 45 degrees

To find the coordinates of the point on the unit circle at an angle of $45^\circ$, we use the fact that the unit circle has a radius of 1.

The coordinates for an angle $\theta$ in radians can be given by $(\cos \theta, \sin \theta)$.

Converting $45^\circ$ to radians:

$$\theta = 45^\circ = \frac{45 \pi}{180} = \frac{\pi}{4}$$

Therefore, the coordinates are:

$$ (\cos \frac{\pi}{4}, \sin \frac{\pi}{4}) $$

Since $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, the coordinates are:

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

Find the tangent of an angle on the unit circle

To find $ \tan(\theta) $ where $ \theta = \frac{5\pi}{4} $:

First, recognize that $ \frac{5\pi}{4} $ is in the third quadrant of the unit circle.

In the third quadrant, the tangent function is positive.

The reference angle for $ \frac{5\pi}{4} $ is:

$$ \pi – \frac{5\pi}{4} = \frac{\pi}{4} $$

Using the reference angle, we have:

$$ \tan(\frac{\pi}{4}) = 1 $$

Thus, $$ \tan(\frac{5\pi}{4}) = 1 $$

Solve for the angle θ in the unit circle where sin(θ) = √3/2 and θ is in the interval [0, 2π) Prove your answer

To solve for $\theta$ in the equation $\sin(\theta) = \frac{\sqrt{3}}{2}$ on the unit circle, we need to determine the angles corresponding to this sine value. The reference angle for $\sin(\theta) = \frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$. Since sine is positive in the first and second quadrants:

1. In the first quadrant, $\theta = \frac{\pi}{3}$

2. In the second quadrant, $\theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3}$

Thus, the solutions are:

$$\theta = \frac{\pi}{3}, \frac{2\pi}{3}$$

Find the equations of the tangents to the unit circle at the point (1/sqrt(2), 1/sqrt(2))

To find the equation of the tangent to the unit circle at a given point, we can use the formula for the tangent line to a circle:

$$x_1 x + y_1 y = 1$$

where $(x_1, y_1)$ is the point of tangency. Here, $x_1 = \frac{1}{\sqrt{2}}$ and $y_1 = \frac{1}{\sqrt{2}}$. Substituting these values into the formula, we get:

$$\frac{1}{\sqrt{2}} x + \frac{1}{\sqrt{2}} y = 1$$

Multiplying both sides by $\sqrt{2}$ to simplify, we obtain:

$$x + y = \sqrt{2}$$

Therefore, the equation of the tangent is:

$$x + y = \sqrt{2}$$

Find the value of cotangent for a given angle on the unit circle

Given an angle $\theta = \frac{5\pi}{6}$, we need to find the value of $\cot \theta$.

The coordinates of the point corresponding to $\theta = \frac{5\pi}{6}$ on the unit circle are $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.

From the definition of cotangent, $\cot \theta = \frac{\cos \theta}{\sin \theta}$.

Here, $\cos \theta = -\frac{\sqrt{3}}{2}$ and $\sin \theta = \frac{1}{2}$.

Therefore,

$$\cot \theta = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$

$$ = -\sqrt{3}$$

So, $\cot \left(\frac{5\pi}{6}\right) = -\sqrt{3}$.

Determine the angles θ in the interval [0, 2π) for which tan(θ) = sqrt(3) and express the solutions in terms of π

Consider the unit circle and the given equation:

$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \sqrt{3}$$

We know that $$\tan(\theta) = \sqrt{3}$$ when $$\theta = \frac{\pi}{3} + k\pi$$, where $$k$$ is an integer. In the interval $$[0, 2\pi)$$, we thus find:

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

Therefore, the solutions are:

$$\boxed{\frac{\pi}{3}, \frac{4\pi}{3}}$$

Calculate the coordinates of the point on the unit circle corresponding to an angle of 5π/6 radians

To find the coordinates of the point on the unit circle corresponding to the given angle, we use the cosine and sine functions:

$$x = \cos\left(\frac{5\pi}{6}\right)$$

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

Since $\frac{5\pi}{6}$ is in the second quadrant, where cosine is negative and sine is positive, we have:

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

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

Therefore, the coordinates are:

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

Find the sine, cosine, and tangent values of the given angles on the unit circle

$$ \text{Consider the angle } \theta = \frac{7\pi}{6} $$

$$ \text{Step 1: Find the sine value } \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} $$

$$ \text{Step 2: Find the cosine value } \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} $$

$$ \text{Step 3: Calculate tangent } \tan\left(\frac{7\pi}{6}\right) = \frac{\sin\left(\frac{7\pi}{6}\right)}{\cos\left(\frac{7\pi}{6}\right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

$$ \text{Answer: } \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}, \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \tan\left(\frac{7\pi}{6}\right) = \frac{\sqrt{3}}{3} $$

Find the value of tan(θ) given specific conditions on the unit circle

Given that $\theta$ is a point on the unit circle where the coordinates are $( \cos(\theta), \sin(\theta) )$ and $\theta$ lies in the second quadrant, find $\tan(\theta)$.

Since $\theta$ is in the second quadrant, $\cos(\theta)$ is negative and $\sin(\theta)$ is positive. Assume $\cos(\theta) = -3/5$, we use the Pythagorean identity to find $\sin(\theta)$:

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

$$(-3/5)^2 + \sin^2(\theta) = 1$$

$$9/25 + \sin^2(\theta) = 1$$

$$\sin^2(\theta) = 1 – 9/25$$

$$\sin^2(\theta) = 16/25$$

Since $\sin(\theta)$ is positive in the second quadrant:

$$\sin(\theta) = 4/5$$

Now, we can find $\tan(\theta)$:

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