Find the angles on the unit circle
Given a point on the unit circle at coordinates (1/2, √3/2), find the corresponding angle in degrees.
The point (1/2, √3/2) corresponds to an angle of 60 degrees.
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Find the angles on the unit circle
Given a point on the unit circle at coordinates (1/2, √3/2), find the corresponding angle in degrees.
The point (1/2, √3/2) corresponds to an angle of 60 degrees.
Find the angle θ in radians for a point on the unit circle that satisfies given conditions
Given a point $ P $ on the unit circle, where the coordinates of $ P $ are $ ( \cos(\theta), \sin(\theta) ) $.
If the coordinates of $ P $ are given as $ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $, we need to determine the angle $ \theta $.
On the unit circle, these coordinates correspond to:
$$ \cos(\theta) = \frac{1}{2} \quad \text{and} \quad \sin(\theta) = \frac{\sqrt{3}}{2} $$
From the unit circle, we know that:
$$ \theta = \frac{\pi}{3} $$
Since the angle $ \theta $ can also be in the second quadrant, we have:
$$ \theta = \frac{5\pi}{3} $$
Find the value of sin(θ), cos(θ), and tan(θ) for θ = π/3 on the unit circle
When $θ = \fracπ3$, we can find the values of $\sin(θ)$, $\cos(θ)$, and $\tan(θ)$ from the unit circle:
$$\sin(\fracπ3) = \frac{\sqrt3}2$$
$$\cos(\fracπ3) = \frac12$$
$$\tan(\fracπ3) = \frac{\sin(\fracπ3)}{\cos(\fracπ3)} = \sqrt3$$
Evaluate the integral of cos(2x) from 0 to pi/2
To evaluate the integral of $ \cos(2x) $ from $ 0 $ to $ \frac{\pi}{2} $:
$$ \int_0^{\frac{\pi}{2}} \cos(2x) \, dx $$
Use the substitution $ u = 2x $, then $ du = 2dx $ or $ dx = \frac{1}{2} du $:
$$ \int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \frac{1}{2} \int_0^{\pi} \cos(u) \, du $$
The integral of $ \cos(u) $ is $ \sin(u) $:
$$ \frac{1}{2} \left[ \sin(u) \right]_0^{\pi} $$
Evaluate the definite integral:
$$ \frac{1}{2} \left( \sin(\pi) – \sin(0) \right) = \frac{1}{2} (0 – 0) = 0 $$
Therefore, the final answer is:
$$ 0 $$
Determine the coordinates of the point on the unit circle corresponding to a given angle
To determine the coordinates of the point on the unit circle corresponding to the angle $\theta$, we use the following formulas for the unit circle:
$$ x = \cos(\theta) $$
$$ y = \sin(\theta) $$
For instance, if $\theta = \frac{\pi}{4}$, then:
$$ x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
$$ y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $$
So, the coordinates are:
$$ \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) $$
Identify the coordinates of the point on the unit circle at an angle of π/4
On the unit circle, the coordinates of the point at an angle of $ \frac{\pi}{4} $ are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
Explain the coordinates of a point on the unit circle at an angle of π/4
The unit circle is a circle with a radius of 1, centered at the origin (0, 0) in the coordinate plane. The coordinates of a point on the unit circle corresponding to an angle of $ \frac{\pi}{4} $ radians can be found using trigonometric functions.
At an angle of $ \frac{\pi}{4} $ radians, the x-coordinate and y-coordinate of the point are:
$$x = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$
$$y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$
So the coordinates are:
$$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$$
Determine the exact values of tan(θ) for θ = 5π/6, θ = 3π/4, and θ = 7π/4 from the unit circle
To determine the exact values of $ \tan(\theta) $ for the given angles using the unit circle, we need to recall the tangent function and its relation to sine and cosine:
\n
$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$
\n
1. For $ \theta = \frac{5\pi}{6} $:
\n
$$ \sin(\frac{5\pi}{6}) = \frac{1}{2}, \quad \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} $$
\n
Therefore:
\n
$$ \tan(\frac{5\pi}{6}) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$
\n
2. For $ \theta = \frac{3\pi}{4} $:
\n
$$ \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}, \quad \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} $$
\n
Therefore:
\n
$$ \tan(\frac{3\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$
\n
3. For $ \theta = \frac{7\pi}{4} $:
\n
$$ \sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2}, \quad \cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2} $$
\n
Therefore:
\n
$$ \tan(\frac{7\pi}{4}) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$
Determine the quadrant in which an angle lies given its sine and cosine values on the unit circle
Given that the sine and cosine values of an angle are both positive, the angle lies in the first quadrant.
Determine the coordinates of a point on the unit circle where the angle θ equals π/4
To determine the coordinates of a point on the unit circle where $ \theta $ equals $ \frac{\pi}{4} $, we use the unit circle equation:
$$ x^2 + y^2 = 1 $$
For $ \theta = \frac{\pi}{4} $, the coordinates are:
$$ \left( \cos \frac{\pi}{4}, \sin \frac{\pi}{4} \right) $$
The values are:
$$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$
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