Unit Circle

Find the tangent of the angle where the unit circle intersects the x-axis at (1, 0)

To find the tangent of the angle, we first note that the point of intersection with the x-axis at (1, 0) corresponds to 0 radians or 0 degrees.

The tangent of an angle in a unit circle is given by $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$.

For $$\theta = 0$$:

$$\sin(0) = 0$$ and $$\cos(0) = 1$$.

Therefore,

$$\tan(0) = \frac{0}{1} = 0$$.

So, the tangent of the angle is 0.

Find the points where the line y = mx + c intersects the unit circle

To find the points of intersection between the line $ y = mx + c $ and the unit circle $ x^2 + y^2 = 1 $, we substitute the expression for y into the circle’s equation:

$$ x^2 + (mx + c)^2 = 1 $$

Expanding and simplifying:

$$ x^2 + m^2x^2 + 2mcx + c^2 = 1 $$

Combining like terms:

$$ (1 + m^2)x^2 + 2mcx + c^2 – 1 = 0 $$

This is a quadratic equation in x. To solve for x, we use the quadratic formula:

$$ x = \frac{-2mc \pm \sqrt{(2mc)^2 – 4(1+m^2)(c^2 – 1)}}{2(1+m^2)} $$

Simplifying under the square root and the denominator:

$$ x = \frac{-2mc \pm \sqrt{4m^2c^2 – 4(1+m^2)(c^2 – 1)}}{2(1+m^2)} $$

$$ x = \frac{-mc \pm \sqrt{m^2c^2 – (1+m^2)(c^2 – 1)}}{1+m^2} $$

$$ x = \frac{-mc \pm \sqrt{m^2c^2 – c^2 – m^2c^2 + m^2 + 1}}{1+m^2} $$

$$ x = \frac{-mc \pm \sqrt{m^2 + 1 – c^2}}{1+m^2} $$

Thus, we find the x-coordinates of the intersection points as:

$$ x_1 = \frac{-mc + \sqrt{m^2 + 1 – c^2}}{1 + m^2}, x_2 = \frac{-mc – \sqrt{m^2 + 1 – c^2}}{1 + m^2} $$

The corresponding y-coordinates are found by substituting these x-values back into the line equation $ y = mx + c $.

Find the sine and cosine of π/4 using the unit circle

To find the sine and cosine of $ \frac{\pi}{4} $ using the unit circle, we can use the coordinates of the corresponding point on the unit circle. For an angle of $ \frac{\pi}{4} $ radians, the coordinates are:

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

Therefore:

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

$$ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

What is the cosine of an angle in the unit circle corresponding to 7π/6 radians?

To find the cosine of the angle $ \frac{7\pi}{6} $ in the unit circle, we first recognize that this angle is in the third quadrant. An angle in the third quadrant has a negative cosine value.

The reference angle for $ \frac{7\pi}{6} $ is $ \frac{\pi}{6} $.

Since the cosine of $ \frac{\pi}{6} $ is $ \frac{\sqrt{3}}{2} $, the cosine of $ \frac{7\pi}{6} $ is $ -\frac{\sqrt{3}}{2} $.

Therefore, $ \cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2} $.

Find the coordinates of the point where the terminal side of an angle in standard position intersects the unit circle if the angle is given by theta = 7π/4

The unit circle has a radius of 1 and is centered at the origin. The coordinates (x, y) on the unit circle for an angle \( \theta \) are given by:

$$ (x, y) = (\cos(\theta), \sin(\theta)) $$

For \( \theta = \frac{7\pi}{4} \):

$$ \cos \left( \frac{7\pi}{4} \right) = \cos \left( 2\pi – \frac{\pi}{4} \right) = \cos \left( \frac{-\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

$$ \sin \left( \frac{7\pi}{4} \right) = \sin \left( 2\pi – \frac{\pi}{4} \right) = \sin \left( \frac{-\pi}{4} \right) = -\sin \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

Therefore, the coordinates are:

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

Given a point on the unit circle (a, b), find the value of cos(θ) and sin(θ)

Given a point on the unit circle $(a, b)$, we can find $\cos(\theta)$ and $\sin(\theta)$:

The coordinates of the point on the unit circle, $(a, b)$, represent the values of $\cos(\theta)$ and $\sin(\theta)$, respectively.

Thus,

$$ \cos(\theta) = a $$

$$ \sin(\theta) = b $$

What is the tangent value at angle pi/4 on the unit circle?

The tangent of an angle $ \theta $ in the unit circle is defined as the ratio of the y-coordinate to the x-coordinate of the corresponding point on the unit circle. For $ \theta = \frac{\pi}{4} $, the coordinates are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

Thus:

$$ \tan\left( \frac{\pi}{4} \right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

Determine the points on the unit circle corresponding to multiples of π/4 and explain their significance in a unit circle art project

To determine the points on the unit circle for multiples of $ \frac{π}{4} $, we first note that:

$$ \theta = n \cdot \frac{π}{4} $$

where $ n $ is an integer. Evaluating this for $ n = 0, 1, 2, 3, 4, 5, 6, 7 $, we get the following points on the unit circle:

– For $ n = 0 $: $ (\cos(0), \sin(0)) = (1, 0) $

– For $ n = 1 $: $ (\cos(\frac{π}{4}), \sin(\frac{π}{4})) = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $

– For $ n = 2 $: $ (\cos(\frac{π}{2}), \sin(\frac{π}{2})) = (0, 1) $

– For $ n = 3 $: $ (\cos(\frac{3π}{4}), \sin(\frac{3π}{4})) = (-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) $

– For $ n = 4 $: $ (\cos(π), \sin(π)) = (-1, 0) $

– For $ n = 5 $: $ (\cos(\frac{5π}{4}), \sin(\frac{5π}{4})) = (-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}) $

– For $ n = 6 $: $ (\cos(\frac{3π}{2}), \sin(\frac{3π}{2})) = (0, -1) $

– For $ n = 7 $: $ (\cos(\frac{7π}{4}), \sin(\frac{7π}{4})) = (\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}) $

These points are significant in a unit circle art project as they help in creating symmetrical designs and patterns based on rotational symmetry.

Calculate the coordinates of a point on the unit circle given the angle θ

To find the coordinates of a point on the unit circle for a given angle $\theta$, use the following formulas:

$$x = \cos(\theta)$$

$$y = \sin(\theta)$$

For example, if $\theta = 45^\circ$:

$$x = \cos(45^\circ) = \frac{\sqrt{2}}{2}$$

$$y = \sin(45^\circ) = \frac{\sqrt{2}}{2}$$

So the coordinates are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$.

Find all solutions for cos(theta) = -1/2 in the range [0, 2pi]

To find all solutions for $ \cos(\theta) = -\frac{1}{2} $ in the range $ [0, 2\pi] $, we need to determine where the cosine function is -1/2 on the unit circle:

Cosine is negative in the second and third quadrants. The reference angle for $ \cos^{-1}(-\frac{1}{2}) $ is $ \frac{\pi}{3} $.

Therefore, the solutions are:

$ \theta_1 = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $

$ \theta_2 = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $

Thus, the solutions are $ \theta = \frac{2\pi}{3} $ and $ \theta = \frac{4\pi}{3} $.