Homework

Find the point on the unit circle where the angle is π/3 and show all steps to verify the trigonometric coordinates

To find the point on the unit circle where the angle is $\frac{\pi}{3}$, we start by noting that the unit circle has a radius of 1. The coordinates of any point on the unit circle can be found using the trigonometric functions cosine (cos) and sine (sin).

We know that for an angle $\theta$:

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

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

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

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

$$ y = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} $$

Therefore, the coordinates of the point on the unit circle where the angle is $\frac{\pi}{3}$ are:

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

Find the coordinates of the point that results from a 225-degree rotation counterclockwise around the origin on the unit circle

To find the coordinates of a point on the unit circle after a $225^\circ$ rotation counterclockwise, we can use the trigonometric functions cosine and sine:

The general formula for finding the coordinates $(x, y)$ on the unit circle is:

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

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

For $\theta = 225^\circ$:

$$x = \cos(225^\circ)$$

$$y = \sin(225^\circ)$$

Since $225^\circ = 180^\circ + 45^\circ$, we can use reference angles:

$$\cos(225^\circ) = \cos(180^\circ + 45^\circ) = -\cos(45^\circ) = -\frac{\sqrt{2}}{2}$$

$$\sin(225^\circ) = \sin(180^\circ + 45^\circ) = -\sin(45^\circ) = -\frac{\sqrt{2}}{2}$$

Therefore, the coordinates are:

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

Find the coordinates of the point on the unit circle corresponding to the angle whose cosine is -2/3

Given the cosine of the angle is $-\frac{2}{3}$,

First, find the sine of the angle using the Pythagorean identity:

$$\cos^2\theta + \sin^2\theta = 1$$

Substitute $\cos\theta = -\frac{2}{3}$:

$$\left(-\frac{2}{3}\right)^2 + \sin^2\theta = 1$$

$$\frac{4}{9} + \sin^2\theta = 1$$

$$\sin^2\theta = 1 – \frac{4}{9}$$

$$\sin^2\theta = \frac{9}{9} – \frac{4}{9}$$

$$\sin^2\theta = \frac{5}{9}$$

Taking the square root,

$$\sin\theta = \pm\sqrt{\frac{5}{9}}$$

$$\sin\theta = \pm\frac{\sqrt{5}}{3}$$

Thus, the coordinates are:

$$(-\frac{2}{3}, \frac{\sqrt{5}}{3})$$ or $$(-\frac{2}{3}, -\frac{\sqrt{5}}{3})$$

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,

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

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