Math

Find the exact coordinates of the point(s) on the unit circle where the tangent line is vertical

The equation of the unit circle is given by:

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

We find the tangent line to be vertical when the derivative is undefined. Thus, we need to find the points where $ \x0crac{dy}{dx} $ is undefined.

Implicitly differentiate the unit circle equation with respect to $ x $:

$$ 2x + 2y \x0crac{dy}{dx} = 0 $$

Simplify and solve for $ \x0crac{dy}{dx} $:

$$ \x0crac{dy}{dx} = -\x0crac{x}{y} $$

The derivative is undefined when $ y = 0 $. Thus, we solve for $ x $:

When $ y = 0 $, substituting back into the original equation:

$$ x^2 = 1 $$

So, $ x = 1 $ or $ x = -1 $.

Therefore, the points are:

$(1,0)$ and $(-1,0)$.

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:

$$\sin(\fracπ3) = \frac{\sqrt3}2$$

$$\cos(\fracπ3) = \frac12$$

$$\tan(\fracπ3) = \frac{\sin(\fracπ3)}{\cos(\fracπ3)} = \sqrt3$$

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:

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

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

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 exact value of sin(π/4) on the unit circle

To find the exact value of $ \sin(\frac{\pi}{4}) $ on the unit circle, we recognize that $ \frac{\pi}{4} $ is equivalent to $ 45^{\circ} $.

On the unit circle, the coordinates for $ \frac{\pi}{4} $ are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

The sine value is the y-coordinate, so:

$$ \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$

Explain the concept of a unit circle, including its importance in trigonometry and how it relates to the coordinates of points on the circle

A unit circle is a circle with a radius of 1, centered at the origin of a coordinate system. The equation of the unit circle is given by:

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

The unit circle is fundamental in trigonometry as it defines the sine and cosine functions for all real numbers. For any angle $\theta$, the coordinates of the corresponding point on the unit circle are $(\cos(\theta), \sin(\theta))$. These coordinates are derived from the definitions:

$$ \cos(\theta) = \frac{x}{1} = x $$

$$ \sin(\theta) = \frac{y}{1} = y $$

Additionally, the unit circle helps in visualizing and understanding periodic properties of trigonometric functions and their symmetries.

Find the angle in degrees corresponding to 7π/6 radians on the unit circle

To convert $\frac{7\pi}{6}$ radians to degrees, we use the conversion factor:

$$ 180^{\circ} = \pi \text{ radians} $$

Thus,

$$ \frac{7\pi}{6} \times \frac{180^{\circ}}{\pi} = 210^{\circ} $$

The angle in degrees is:

$$ 210^{\circ} $$

Find the coordinates on the unit circle where the tangent of the angle is 1

To find the coordinates on the unit circle where $ \tan(\theta) = 1 $, we need to determine the angles $\theta $ for which this condition holds. We know that:

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

For the tangent to be 1, the sine and cosine must be equal. This occurs at angles:

$$ \theta = \frac {\pi}{4} \text{ and } \theta = \frac {5\pi}{4} $$

Now, we find the coordinates on the unit circle for these angles:

$$ \text{At } \theta = \frac {\pi}{4}, \text{ the coordinates are } \left( \frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2} \right) $$

$$ \text{At } \theta = \frac {5\pi}{4}, \text{ the coordinates are } \left( -\frac {\sqrt {2}}{2}, -\frac {\sqrt {2}}{2} \right) $$

Thus, the coordinates on the unit circle where $ \tan(\theta) = 1 $ are:

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

and

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

Determine the coordinates of the point where the terminal side of an angle of 5π/3 radians intersects the unit circle, and identify its quadrant

The angle $ \frac{5\pi}{3} $ radians is equivalent to 300 degrees (since $ \frac{5\pi}{3} \times \frac{180}{\pi} = 300 $ degrees).

This angle places the terminal side in the fourth quadrant.

In the fourth quadrant, the coordinates on the unit circle corresponding to an angle of 300 degrees are:

$$ ( \cos(300\degree), \sin(300\degree) ) $$

Since $ \cos(300\degree) = \cos(-60\degree) = \frac{1}{2} $ and $ \sin(300\degree) = \sin(-60\degree) = -\frac{\sqrt{3}}{2} $, the coordinates are:

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

Thus, the terminal side intersects the unit circle at $ \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) $ in the fourth quadrant.