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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 $\frac{\sqrt{3}}{2}$.
Find the exact values of sin(3π/4), cos(3π/4), and tan(3π/4) using the unit circle
We are asked to find the exact values of $\sin(\frac{3\pi}{4})$, $\cos(\frac{3\pi}{4})$, and $\tan(\frac{3\pi}{4})$ using the unit circle.
First, we locate the angle $\frac{3\pi}{4}$ on the unit circle: it is in the second quadrant.
The reference angle for $\frac{3\pi}{4}$ is $\frac{\pi}{4}$ (45 degrees). In the second quadrant, the sine value is positive, and the cosine value is negative.
Thus, we have:
$$\sin(\frac{3\pi}{4}) = \sin(\pi – \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$
$$\cos(\frac{3\pi}{4}) = \cos(\pi – \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$
$$\tan(\frac{3\pi}{4}) = \frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1$$
Find the angle on the unit circle in the complex plane where the cosine value is 1/2
We start by knowing that the cosine function gives the real part of the point on the unit circle corresponding to a given angle.
We are given $\cos(\theta) = \frac{1}{2}$ and need to find the angles $\theta$ where this holds true.
On the unit circle, $\cos(\theta)$ reaches $\frac{1}{2}$ at two points: $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$.
Therefore, the angles are:
$$\theta = \frac{\pi}{3}, \frac{5\pi}{3}$$
Find the sine and cosine values for the angle θ = 30° on the unit circle
To find the sine and cosine values for $ \theta = 30° $ on the unit circle, we need to locate the point on the unit circle corresponding to $ \theta = 30° $.
1. Convert degrees to radians, $ \theta = 30° = \frac{π}{6} $ radians.
2. From the unit circle, the coordinates for $ \theta = \frac{π}{6} $ are $ ( \cos(30°), \sin(30°) ) $.
3. Hence, $ \cos(30°) = \frac{\sqrt{3}}{2} $ and $ \sin(30°) = \frac{1}{2} $.
So, the cosine value is $ \frac{\sqrt{3}}{2} $ and the sine value is $ \frac{1}{2} $.
Find the value of tan(θ) for θ in the unit circle
To find the value of $\tan(\theta)$ for $\theta$ on the unit circle, consider the point $P(x, y)$ on the circle corresponding to the angle $\theta$. The tangent of $\theta$ is given by the ratio of the y-coordinate to the x-coordinate, i.e., $\tan(\theta) = \frac{y}{x}$. For example, if $\theta = \frac{\pi}{4}$, the coordinates of the corresponding point on the unit circle are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$. Therefore,
$$\tan \left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1.$$
Determine the Quadrant of a Given Point on a Unit Circle
Given the point \((x, y)\) on a unit circle, determine the quadrant in which the point lies.
The unit circle has a radius of 1. The quadrants are defined as follows:
– Quadrant I: \((x > 0, y > 0)\)
– Quadrant II: \((x < 0, y > 0)\)
– Quadrant III: \((x < 0, y < 0)\)
– Quadrant IV: \((x > 0, y < 0)\)
Let’s solve for the point \((-\frac{1}{2}, \frac{\sqrt{3}}{2})\)
Given: \(x = -\frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\)
Since \(x < 0\) and \(y > 0\), the point lies in Quadrant II.
Find the value of tan(θ) at three specific angles on the unit circle: θ = π/4, 3π/4, and 5π/6
For $\theta = \frac{\pi}{4}$:
On the unit circle, at $\theta = \frac{\pi}{4}$, the coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.
The tangent function is given by $\tan(\theta) = \frac{y}{x}$.
Thus,
$$ \tan(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$
For $\theta = \frac{3\pi}{4}$:
On the unit circle, at $\theta = \frac{3\pi}{4}$, the coordinates are $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.
Thus,
$$ \tan(\frac{3\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$
For $\theta = \frac{5\pi}{6}$:
On the unit circle, at $\theta = \frac{5\pi}{6}$, the coordinates are $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.
Thus,
$$ \tan(\frac{5\pi}{6}) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$
Convert the angle 150 degrees to radians and find its coordinates on the unit circle
To convert degrees to radians, we use the conversion factor \( \frac{\pi}{180} \).
$$150^{\circ} \times \frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6}$$
The coordinates on the unit circle for \( \theta = \frac{5\pi}{6} \) are given by \((\cos(\theta), \sin(\theta))\).
$$\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$$
Thus, the coordinates are:
$$( -\frac{\sqrt{3}}{2}, \frac{1}{2} )$$
If $\theta$ is an angle in the unit circle such that $\cos(\theta) = \frac{1}{2}$ and $\sin(\theta) = \frac{\sqrt{3}}{2}$, find the value of $\theta$ in degrees and radians
To solve this problem, we need to determine the angle $\theta$ on the unit circle. Given:
$$\cos(\theta) = \frac{1}{2}$$ $$\sin(\theta) = \frac{\sqrt{3}}{2}$$
On the unit circle, these values correspond to the angle $\theta = 60^{\circ}$ or $\theta = \frac{\pi}{3}$ radians.
Therefore, the value of $\theta$ is:
$$\theta = 60^{\circ}$$ $$\theta = \frac{\pi}{3}$$
Find the sine and cosine values at different points on the unit circle
To find the sine and cosine values at different points on the unit circle, we can use the angle in radians.
1. For the angle $$\frac{\pi}{6}$$ radians:
The coordinates on the unit circle are $$\left( \cos \frac{\pi}{6}, \sin \frac{\pi}{6} \right)$$.
Thus, we get: $$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$ and $$\sin \frac{\pi}{6} = \frac{1}{2}$$.
2. For the angle $$\frac{\pi}{4}$$ radians:
The coordinates on the unit circle are $$\left( \cos \frac{\pi}{4}, \sin \frac{\pi}{4} \right)$$.
Thus, we get: $$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$ and $$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$.
3. For the angle $$\frac{\pi}{3}$$ radians:
The coordinates on the unit circle are $$\left( \cos \frac{\pi}{3}, \sin \frac{\pi}{3} \right)$$.
Thus, we get: $$\cos \frac{\pi}{3} = \frac{1}{2}$$ and $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$.
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