Math

Given a unit circle centered at the origin, but flipped in a non-standard way such that the positive x-axis points downwards and the positive y-axis points to the left, find the coordinates of the point corresponding to an angle of 5π/6 radians

To solve this problem, we first need to understand the transformation of the coordinate system.

In the standard unit circle, an angle of $\frac{5\pi}{6}$ radians would correspond to the point $(-\cos(\frac{\pi}{6}), \sin(\frac{\pi}{6}))$.

Therefore, in the standard unit circle, the coordinates would be:

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

Now, since the unit circle is flipped such that the positive x-axis points downwards and the positive y-axis points to the left, we need to adjust these coordinates accordingly:

1. The x-coordinate will become the negative of the original y-coordinate.

2. The y-coordinate will become the negative of the original x-coordinate.

Thus, the transformed coordinates are:

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

which simplifies to:

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

Find the value of cos(θ) on the unit circle for a given θ and determine the exact coordinates of the corresponding point

Let’s consider the angle $ \theta = \frac{7\pi}{6}$.

First, we determine the reference angle. Since $\frac{7\pi}{6}$ is in the third quadrant, we find the reference angle by subtracting $\pi$:

$$ \theta_{ref} = \frac{7\pi}{6} – \pi = \frac{7\pi}{6} – \frac{6\pi}{6} = \frac{\pi}{6} $$

The cosine of the reference angle $\frac{\pi}{6}$ is $\frac{\sqrt{3}}{2}$, but since we are in the third quadrant, the cosine value is negative:

$$ \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} $$

The exact coordinates of the point on the unit circle corresponding to $\theta = \frac{7\pi}{6}$ are:

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

Find the coordinates of the point on the unit circle where the angle with the positive x-axis is 5π/6 radians

Given the angle $\theta = \frac{5\pi}{6}$ radians, we need to find the coordinates of the point on the unit circle.

On the unit circle, the coordinates of a point at an angle $\theta$ are $$(\cos(\theta), \sin(\theta))$$.

Therefore,

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

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

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

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

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

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

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