Unit Circle

Find the coordinates of the point where the line y = 2x intersects the unit circle

First, let’s write the equation of the unit circle:

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

Since $y = 2x$, we can substitute $2x$ for $y$ in the unit circle equation:

$$x^2 + (2x)^2 = 1.$$

This simplifies to:

$$x^2 + 4x^2 = 1$$

$$5x^2 = 1$$

$$x^2 = \frac{1}{5}$$

$$x = \pm \frac{1}{\sqrt{5}}$$

Substituting these values back into $y = 2x$, we get:

$$y = 2(\frac{1}{\sqrt{5}}) = \frac{2}{\sqrt{5}}$$

$$y = 2(-\frac{1}{\sqrt{5}}) = -\frac{2}{\sqrt{5}}$$

Hence, the points of intersection are:

$$ \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)$$ and $$ \left( -\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}} \right).$$

Find the sine, cosine, and tangent of an angle of 45 degrees on the unit circle

To find the sine, cosine, and tangent of an angle of 45 degrees on the unit circle:

The coordinates of the point at $45^\circ$ on the unit circle are $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.
Therefore, $\sin(45^\circ) = \frac{\sqrt{2}}{2}$ and $\cos(45^\circ) = \frac{\sqrt{2}}{2}$.

To find $\tan(45^\circ)$, we use the formula $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.
Thus, $\tan(45^\circ) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$.

Given the unit circle, find the coordinates of the point where the angle θ intersects the unit circle Let θ = 45 degrees

To find the coordinates of the point where the angle $\theta = 45^\circ$ intersects the unit circle, we use the fact that the unit circle has a radius of 1. The coordinates on the unit circle are given by $(\cos \theta, \sin \theta)$.

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

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

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

Determine the Quadrants of Trigonometric Values on the Unit Circle

To determine the quadrant of the angle $\frac{5\pi}{3}$ on the unit circle:

1. Identify the reference angle: $\frac{5\pi}{3} – 2\pi = \frac{-\pi}{3}$, which is equal to $\frac{\pi}{3}$.

2. Determine the quadrant where $\frac{5\pi}{3}$ lies:

$\frac{5\pi}{3}$ is between $\frac{3\pi}{2}$ and $2\pi$, so it lies in the fourth quadrant.

The answer is Quadrant IV.

Find the cosine values of the angles on the unit circle

Given the angle $\theta = \frac{5\pi}{3}$, we need to find the cosine value.

The unit circle coordinates at an angle $\theta$ are given by $(\cos(\theta), \sin(\theta))$. For $\theta = \frac{5\pi}{3}$, the angle is in the fourth quadrant where the cosine is positive and sine is negative.

Using reference angles, we can see that $\frac{5\pi}{3}$ is equivalent to $-\frac{\pi}{3}$ or $2\pi – \frac{\pi}{3}$. Thus, the cosine value is:

$$\cos\left(\frac{5\pi}{3}\right) = \cos\left(2\pi – \frac{\pi}{3}\right) = \cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right)$$

From the unit circle, we know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. Therefore,

$$\cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}$$

Find the points of intersection between the unit circle and the line y = 2x + 1

To find the points of intersection, we can substitute $y = 2x + 1$ into the equation of the unit circle, which is $x^2 + y^2 = 1$.

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

Expanding the equation:

$$x^2 + (4x^2 + 4x + 1) = 1$$

Combining like terms:

$$5x^2 + 4x + 1 = 1$$

Simplifying:

$$5x^2 + 4x = 0$$

Factoring the equation:

$$x(5x + 4) = 0$$

So $x = 0$ or $x = -\frac{4}{5}$.

When $x = 0$, $y = 1$.

When $x = -\frac{4}{5}$, $y = 2(-\frac{4}{5}) + 1 = -\frac{8}{5} + 1 = -\frac{3}{5}$.

Thus, the points of intersection are $(0, 1)$ and $(-\frac{4}{5}, -\frac{3}{5})$.

Find the angle θ in the unit circle where cos(θ) = 05

$$\text{Given } \cos(\theta) = 0.5$$

$$\text{We know that } \cos(\theta) = 0.5 \text{ at } \theta = \frac{\pi}{3} \text{ and } \theta = -\frac{\pi}{3} \text{ (or equivalently } \theta = 2\pi – \frac{\pi}{3} = \frac{5\pi}{3} \text{)}$$

$$\text{Therefore, the angles } \theta \text{ in radians where } \cos(\theta) = 0.5 \text{ are } \theta = \frac{\pi}{3} \text{ and } \theta = \frac{5\pi}{3}.$$

Find the exact values of the sine and cosine of an angle using the unit circle

To find the exact values of $\sin(\theta)$ and $\cos(\theta)$ using the unit circle, let’s consider $\theta = \frac{5\pi}{6}$.

First, we know that $\frac{5\pi}{6}$ is in the second quadrant.

In the second quadrant, sine is positive and cosine is negative.

Using the reference angle $\frac{\pi}{6}$, we have:

$$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi – \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$

and

$$\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\pi – \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$

Determine the expression for the flipped unit circle equation

To determine the equation for a unit circle flipped over the y-axis, we start with the standard unit circle equation:

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

When we flip the unit circle over the y-axis, we change the sign of the x-coordinate. Therefore, the new equation becomes:

$$(-x)^2 + y^2 = 1$$

Simplifying this, we get:

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

Thus, the equation of the unit circle flipped over the y-axis is the same as the original unit circle.

Find the value of cos θ on the unit circle in the complex plane when θ = π/3

To find the value of $\cos \theta$ on the unit circle, we use the unit circle definition where the coordinates are $(\cos \theta, \sin \theta)$.

For $\theta = \pi/3$, the coordinates on the unit circle are:

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

Therefore,

$$\cos \frac{\pi}{3} = \frac{1}{2}$$