Unit Circle

On the unit circle, find the value of cos(135°) + sin(225°) + tan(315°)

$$ \cos(135°) $$

Since $135°$ lies in the second quadrant, we have:

$$ \cos(135°) = -\cos(180° – 135°) = -\cos(45°) = -\frac{\sqrt{2}}{2} $$

$$ \sin(225°) $$

Since $225°$ lies in the third quadrant, we have:

$$ \sin(225°) = -\sin(360° – 225°) = -\sin(135°) = -\sin(180° – 135°) = -\sin(45°) = -\frac{\sqrt{2}}{2} $$

$$ \tan(315°) $$

Since $315°$ lies in the fourth quadrant, we have:

$$ \tan(315°) = \tan(360° – 45°) = \tan(45°) = 1 $$

Combining all these, we get:

$$ \cos(135°) + \sin(225°) + \tan(315°) = -\frac{\sqrt{2}}{2} + -\frac{\sqrt{2}}{2} + 1 = -\sqrt{2} + 1 $$

If the standard unit circle is flipped over the x-axis, describe the transformation of the angle θ and calculate the new coordinates for θ = 2π/3

When the unit circle is flipped over the x-axis, the y-coordinates of all points on the circle are inverted. Therefore, the angle $\theta$ remains the same in magnitude but the y-value of the coordinate changes sign.

For $\theta = \frac{2\pi}{3}$, the original coordinates on the unit circle are:

$$\left(\cos\left(\frac{2\pi}{3}\right), \sin\left(\frac{2\pi}{3}\right)\right)$$

We compute:

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

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

Therefore, the original coordinates are:
$$\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$

After flipping over the x-axis, the new coordinates become:
$$\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$$

Find the angle θ in the unit circle such that cos(θ) = -1/2

We know that $ \cos(\theta) = -\frac{1}{2} $.

This value of cosine corresponds to two angles in the unit circle, which are in the second and third quadrants.

In the second quadrant, the reference angle is $ \theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $.

In the third quadrant, the reference angle is $ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $.

Therefore, $ \theta = \frac{2\pi}{3} $ or $ \theta = \frac{4\pi}{3} $.

Suppose that angle θ is positioned on the unit circle such that θ = 5π/6 Determine the coordinates of the point where the terminal side of θ intersects the unit circle

First, we identify that $ \theta = \frac{5\pi}{6} $ is in the second quadrant. The reference angle is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.

In the unit circle, the cosine and sine of $ \frac{\pi}{6} $ are $ \frac{\sqrt{3}}{2} $ and $ \frac{1}{2} $, respectively.

Therefore, in the second quadrant, the coordinates are $ (-\frac{\sqrt{3}}{2}, \frac{1}{2}) $.

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

How to Find the Reference Angle Not on Unit Circle

To find the reference angle of an angle not on the unit circle, follow these steps:

1. Determine the quadrant in which the angle is located.

2. Use the following rules based on the quadrant to find the reference angle:

For an angle $\theta$ in the first quadrant, the reference angle is $\theta$.

For an angle $\theta$ in the second quadrant, the reference angle is $180^\circ – \theta$.

For an angle $\theta$ in the third quadrant, the reference angle is $\theta – 180^\circ$.

For an angle $\theta$ in the fourth quadrant, the reference angle is $360^\circ – \theta$.

Example: Find the reference angle for $210^\circ$.

Since $210^\circ$ is in the third quadrant, we use the rule for the third quadrant:

$$\text{Reference Angle} = 210^\circ – 180^\circ = 30^\circ$$

Therefore, the reference angle for $210^\circ$ is $30^\circ$.

Find the coordinates of a point on the unit circle at a given angle

To find the coordinates of a point on the unit circle at an angle of 120 degrees, we first convert the angle to radians, as the unit circle typically uses radians.

$$ \theta = 120^{\circ} = \frac{2\pi}{3} \text{ radians} $$

The coordinates on the unit circle are given by:

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

Substituting our angle:

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

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

Thus, the coordinates of the point on the unit circle at 120 degrees are:

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

Find the coordinates of the point on the unit circle at angle π/3 radians

To find the coordinates of the point on the unit circle at angle $\pi/3$ radians, we use the unit circle definitions. The unit circle is defined by the equation $x^2 + y^2 = 1$, where the coordinates $(x, y)$ correspond to $(\cos(\theta), \sin(\theta))$ for an angle $\theta$.

For $\theta = \pi/3$:

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

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

Thus, the coordinates are $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$.

Find the intersection points of the unit circle and the line y = x – 1

To find the intersection points, we need to solve the system of equations formed by the unit circle equation and the given line equation.

The unit circle equation is:

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

Substitute $y = x – 1$ into the unit circle equation:

$$x^2 + (x – 1)^2 = 1$$

Expand and simplify:

$$x^2 + x^2 – 2x + 1 = 1$$

Combine like terms:

$$2x^2 – 2x + 1 = 1$$

Simplify further:

$$2x^2 – 2x = 0$$

Factor out the common term:

$$2x(x – 1) = 0$$

Set each factor to zero:

$$x = 0$$

$$x = 1$$

For $x = 0$:

$$y = 0 – 1 = -1$$

For $x = 1$:

$$y = 1 – 1 = 0$$

Thus, the intersection points are $(0, -1)$ and $(1, 0)$.

Given the unit circle, if the point (a, b) lies on the circle, find the value of sin(2θ), cos(2θ), and tan(2θ) where θ is the angle that corresponds to the point (a, b) Verify that these values satisfy the double angle identities

Given the point (a, b) on the unit circle, we know:

$$a = \cos \theta $$

$$b = \sin \theta $$

We need to find $\sin(2\theta)$, $\cos(2\theta)$, and $\tan(2\theta)$. Using the double angle identities:

$$ \sin(2\theta) = 2 \sin \theta \cos \theta $$

$$ \cos(2\theta) = \cos^2 \theta – \sin^2 \theta $$

$$ \tan(2\theta) = \frac{2\tan \theta}{1 – \tan^2 \theta} $$

By substituting a = cos θ and b = sin θ:

$$ \sin(2\theta) = 2ab $$

$$ \cos(2\theta) = a^2 – b^2 $$

$$ \tan(2\theta) = \frac{2b/a}{1 – b^2/a^2} = \frac{2b/a}{(a^2 – b^2)/a^2} = \frac{2ab}{a^2 – b^2} $$

Verification of double-angle identities:

$$ (2ab)^2 + (a^2 – b^2)^2 = 4a^2b^2 + a^4 – 2a^2b^2 + b^4 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2 = 1 $$

Find the coordinates of point P on the unit circle at an angle of 45 degrees

To find the coordinates of point P on the unit circle at an angle of 45 degrees, we use the fact that the unit circle has a radius of 1 and that the coordinates correspond to the cosine and sine of the angle.

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

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

Therefore, the coordinates of point P are:

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