Explore the unit circle and its relationship to angles, radians, trigonometric ratios, and coordinates in the coordinate plane.
Find the value of sin(π/3) on the unit circle
To find the value of $ \sin\left(\frac{\pi}{3}\right) $ on the unit circle, we look at the reference angle for $ \frac{\pi}{3} $.
The reference angle is $ 60^{\circ} $, and the sine value for $ 60^{\circ} $ is:
$$ \sin\left(60^{\circ}\right) = \frac{\sqrt{3}}{2} $$
Therefore:
$$ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} $$
Find the area of a sector of a unit circle with a central angle of θ radians
To find the area of a sector of a unit circle with a central angle of $ \theta $, we use the formula for the area of a sector:
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$$ A = \frac{1}{2} r^2 \theta $$
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Since the radius $ r $ of a unit circle is 1, the formula simplifies to:
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$$ A = \frac{1}{2} \cdot 1^2 \cdot \theta = \frac{1}{2} \theta $$
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The area of the sector is:
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$$ A = \frac{\theta}{2} $$
Find the value of sin(π/6), cos(π/3), and tan(π/4) using the unit circle chart
Using the unit circle chart, we find:
$$ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} $$
$$ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} $$
$$ \tan\left(\frac{\pi}{4}\right) = 1 $$
Find the value of cos(π/4) and sin(π/4) using the unit circle
To find the values of $ \cos(\frac{\pi}{4}) $ and $ \sin(\frac{\pi}{4}) $ using the unit circle, we need to identify the coordinate point on the unit circle that corresponds to the angle $ \frac{\pi}{4} $.
The angle $ \frac{\pi}{4} $ is located in the first quadrant where both sine and cosine values are positive. This angle corresponds to the point $ (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $ on the unit circle.
Therefore:
$$ \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$
$$ \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$
Find the equation of the tangent line to the unit circle at (1, 0)
The unit circle is given by the equation:
$$ x^2 + y^2 = 1 $$
To find the equation of the tangent line at $(1, 0)$, we first find the slope of the tangent. Differentiate the equation implicitly with respect to $x$:
$$ 2x + 2y \x0crac{dy}{dx} = 0 $$
At the point $(1, 0)$, substitute $x = 1$ and $y = 0$:
$$ 2(1) + 2(0) \x0crac{dy}{dx} = 0 $$
So, the slope $\x0crac{dy}{dx}$ at $(1, 0)$ is $0$. The equation of the tangent line using point-slope form is:
$$ y – 0 = 0(x – 1) $$
Therefore, the equation is:
$$ y = 0 $$
What does the sine function represent on the unit circle?
On the unit circle, the sine function $ \sin(\theta) $ represents the y-coordinate of a point on the circle corresponding to the angle $ \theta $ measured from the positive x-axis.
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Mathematically, if a point on the unit circle is given by $ (x, y) $, then for any angle $ \theta $, the coordinates can be expressed as:
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$$ x = \cos(\theta) $$
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$$ y = \sin(\theta) $$
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This implies that the sine of an angle is the vertical distance from the x-axis to the point on the circle.
Determine the values of trigonometric functions for specific angles on the unit circle
Given the angle $ \theta = \frac{5\pi}{4} $ radians, find the values of $ \sin(\theta) $, $ \cos(\theta) $, and $ \tan(\theta) $.
The coordinates of $ \frac{5\pi}{4} $ on the unit circle are $ \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right) $.
So, $ \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} $, $ \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} $, and
$$ \tan\left(\frac{5\pi}{4}\right) = \frac{\sin\left(\frac{5\pi}{4}\right)}{\cos\left(\frac{5\pi}{4}\right)} = 1. $$
Find the value of arcsin(1/2) using the unit circle
To find the value of $ \arcsin(\frac{1}{2}) $ using the unit circle, we need to determine the angle whose sine is $ \frac{1}{2} $.
On the unit circle, the sine of an angle is the y-coordinate of the corresponding point.
The angle that has a sine of $ \frac{1}{2} $ in the range $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $ is $ \frac{\pi}{6} $.
Therefore, the value of $ \arcsin(\frac{1}{2}) $ is:
$$ \frac{\pi}{6} $$
Determine the coordinates of a point on the unit circle where the sine value is 1/2 and the tangent value is positive
To find the coordinates where $\sin(\theta) = \frac{1}{2}$ and $\tan(\theta)$ is positive, we analyze the unit circle.
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The sine function equals $\frac{1}{2}$ at two angles: $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.
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Since the tangent function is positive when both sine and cosine have the same sign, we consider the angles in the first and third quadrants.
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For $\theta = \frac{\pi}{6}$, the coordinates on the unit circle are:
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$$ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) = (\frac{\sqrt{3}}{2}, \frac{1}{2}) $$
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Thus, the coordinates are:
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$$ (\frac{\sqrt{3}}{2}, \frac{1}{2}) $$
Find the value of tan(θ) using the unit circle when θ is in the third quadrant
To find the value of $ \tan(θ) $ using the unit circle, we need to determine the coordinates where $ θ $ intersects the unit circle in the third quadrant.
In the third quadrant, both the x and y coordinates are negative. Suppose $ θ = 225° $ (or $ \frac{5π}{4} $ in radians). In this case, the coordinates on the unit circle are $ ( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} ) $.
The tangent of $ θ $ is given by the ratio of the y-coordinate to the x-coordinate:
$$ \tan(225°) = \frac{y}{x} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 $$
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