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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 $\tan(\theta)$:
$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{4/5}{-3/5} = -\frac{4}{3}$$
Find the angles on the unit circle where cos(θ) = -1/2
To find the angles where $\cos(\theta) = -\frac{1}{2}$ on the unit circle, we need to consider the unit circle properties and the cosine function.
1. The cosine of an angle represents the x-coordinate on the unit circle.
2. $\cos(\theta) = -\frac{1}{2}$ corresponds to the x-coordinate -1/2.
3. The angles with $\cos(\theta) = -\frac{1}{2}$ are in the second and third quadrants because cosine is negative in these quadrants.
4. The reference angle for $\cos(\theta) = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$.
5. Therefore, the angles are:
$\theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3}$
$\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
Thus, the angles are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.
What is the cosine of 30 degrees on the unit circle?
To find the cosine of $30^{\circ}$ on the unit circle, we need to recall the coordinates of the point on the unit circle corresponding to $30^{\circ}$. The coordinates are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$.
The cosine of an angle is given by the x-coordinate of the corresponding point on the unit circle.
Therefore, the cosine of $30^{\circ}$ is $\frac{\sqrt{3}}{2}$.
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