Unit Circle

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:

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

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1. For $ \theta = \frac{5\pi}{6} $:

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$$ \sin(\frac{5\pi}{6}) = \frac{1}{2}, \quad \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} $$

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Therefore:

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$$ \tan(\frac{5\pi}{6}) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$

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2. For $ \theta = \frac{3\pi}{4} $:

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$$ \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}, \quad \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} $$

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Therefore:

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$$ \tan(\frac{3\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

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3. For $ \theta = \frac{7\pi}{4} $:

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$$ \sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2}, \quad \cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2} $$

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Therefore:

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$$ \tan(\frac{7\pi}{4}) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$

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:

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

Find the value of sec(θ) at θ = π/3 on the unit circle

To find the value of $ \sec(θ) $ at $ θ = \frac{\pi}{3} $ on the unit circle, we first find the cosine of the angle:

$$ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} $$

Then, since $ \sec(θ) $ is the reciprocal of $ \cos(θ) $:

$$ \sec\left(\frac{\pi}{3}\right) = \frac{1}{\cos\left(\frac{\pi}{3}\right)} = \frac{1}{\frac{1}{2}} = 2 $$

Identify the coordinates of points on the unit circle for given angles

For the angle $ \theta = \frac{\pi}{6} $, the point on the unit circle is given by $ (\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6})) $.

Calculate these values:

$$ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $$

$$ \sin(\frac{\pi}{6}) = \frac{1}{2} $$

Therefore, the coordinates are:

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

Find the equation of a tangent to the unit circle at a given point

To find the equation of a tangent to the unit circle at the point $(a, b)$, we start by noting that the unit circle is defined by:

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

The slope of the radius at $(a, b)$ is $ \x0crac{b}{a} $, so the slope of the tangent line, being perpendicular to the radius, is:

$$ -\x0crac{a}{b} $$

Using the point-slope form of a line, the equation of the tangent line can be written as:

$$ y – b = -\x0crac{a}{b}(x – a) $$

Simplifying, we get:

$$ bx + ay = 1 $$

Find the values of tan(θ) for θ in the unit circle at 0, π/4, π/3, and π/2

To determine the values of $ \tan(\theta) $ for $ \theta $ in the unit circle at $ 0 $, $ \frac{\pi}{4} $, $ \frac{\pi}{3} $, and $ \frac{\pi}{2} $, we evaluate the tangent function at these angles:

For $ \theta = 0 $:

$$ \tan(0) = 0 $$

For $ \theta = \frac{\pi}{4} $:

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

For $ \theta = \frac{\pi}{3} $:

$$ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} $$

For $ \theta = \frac{\pi}{2} $:

$$ \tan\left(\frac{\pi}{2}\right) = \text{undefined} $$

Find the angle $ \theta $ on the unit circle where the following conditions are met: $ \sin(\theta) = -\frac{1}{2} $ and $ \cos(\theta) = -\frac{\sqrt{3}}{2} $

To find the angle $ \theta $ on the unit circle where $ \sin(\theta) = -\frac{1}{2} $ and $ \cos(\theta) = -\frac{\sqrt{3}}{2} $, we need to identify the corresponding angles in degrees.

First, note that $ \sin(\theta) = -\frac{1}{2} $ occurs at:

$$ \theta = 210^\circ, 330^\circ $$

Next, note that $ \cos(\theta) = -\frac{\sqrt{3}}{2} $ occurs at:

$$ \theta = 150^\circ, 210^\circ $$

The common angle is:

$$ \theta = 210^\circ $$