Math

Find the Cartesian coordinates of a point on the unit circle where the angle is 135 degrees

To find the Cartesian coordinates of a point on the unit circle where the angle is $135^{\circ}$, we use the unit circle equation:

$$x = \cos(135^{\circ})$$

$$y = \sin(135^{\circ})$$

First, we calculate the cosine and sine of $135^{\circ}$:

$$\cos(135^{\circ}) = -\frac{\sqrt{2}}{2}$$

$$\sin(135^{\circ}) = \frac{\sqrt{2}}{2}$$

So, the Cartesian coordinates are:

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

Find the exact value of sin(2θ) and cos(2θ) given the point on the unit circle and the quadrant

Given the point $ P = \left( -\frac{3}{5}, -\frac{4}{5} \right) $ on the unit circle, find the exact values of $ \sin(2\theta) $ and $ \cos(2\theta) $. The point $ P $ is in Quadrant III.

To find $ \theta $, we use the definitions of sine and cosine on the unit circle:

$$ \sin(\theta) = y = -\frac{4}{5} $$

$$ \cos(\theta) = x = -\frac{3}{5} $$

Using the double angle formulas:

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

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

Substituting the values:

$$ \sin(2\theta) = 2 \left( -\frac{4}{5} \right) \left( -\frac{3}{5} \right) = 2 \times \frac{12}{25} = \frac{24}{25} $$

$$ \cos(2\theta) = \left( -\frac{3}{5} \right)^2 – \left( -\frac{4}{5} \right)^2 = \frac{9}{25} – \frac{16}{25} = -\frac{7}{25} $$

Determine the sine, cosine, and tangent of the angle θ = 30° on the unit circle

To determine the sine, cosine, and tangent of $\theta = 30^\circ$ on the unit circle, we first need to recall the unit circle values for this angle.

For $\theta = 30^\circ$:

$$\sin(30^\circ) = \frac{1}{2}$$

$$\cos(30^\circ) = \frac{\sqrt{3}}{2}$$

$$\tan(30^\circ) = \frac{\sin(30^\circ)}{\cos(30^\circ)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

If the secant of an angle θ in the unit circle is 2, find the angle θ

Given $\sec(\theta) = 2$, we know $\sec(\theta) = \frac{1}{\cos(\theta)}$.

So, $\frac{1}{\cos(\theta)} = 2$ implies $\cos(\theta) = \frac{1}{2}$.

The cosine of $\theta$ is positive, so $\theta$ must be in the first or fourth quadrant.

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

Determine the exact values of trigonometric functions at specific angles on the unit circle

To find the exact values of the trigonometric functions for the angle $\frac{7\pi}{6}$:

1. Find the reference angle by subtracting $\pi$: $$\frac{7\pi}{6} – \pi = \frac{7\pi}{6} – \frac{6\pi}{6} = \frac{\pi}{6}$$

2. Determine the coordinates on the unit circle for $\frac{\pi}{6}$, which are $(\cos \frac{\pi}{6}, \sin \frac{\pi}{6})$. These are $$\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$$

3. Since $\frac{7\pi}{6}$ is in the third quadrant, both cosine and sine are negative: $$\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$$

Thus, $$\cos \left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2}$$ and $$\sin \left( \frac{7\pi}{6} \right) = -\frac{1}{2}$$

Determine the values of sine and cosine at a given angle on the unit circle

Given an angle of $\theta = \frac{5\pi}{6}$, determine the values of $\sin(\theta)$ and $\cos(\theta)$ using the unit circle.

First, recognize that $\theta = \frac{5\pi}{6}$ is located in the second quadrant.

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

The reference angle for $\frac{5\pi}{6}$ is: $\pi – \frac{5\pi}{6} = \frac{\pi}{6}$.

Knowing the values from the unit circle:

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

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

Since $\theta = \frac{5\pi}{6}$ lies in the second quadrant, we have:

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

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

Find the value of sin(θ), cos(θ), and tan(θ) for θ = 5π/6

First, locate $ \theta = \frac{5\pi}{6} $ on the unit circle. This angle is in the second quadrant.

In the second quadrant, the sine function is positive and the cosine function is negative. The reference angle for $ \theta = \frac{5\pi}{6} $ is $ \frac{\pi}{6} $.

The sine and cosine values for $ \frac{\pi}{6} $ are $ \sin(\frac{\pi}{6}) = \frac{1}{2} $ and $ \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} $.

Since $ \theta = \frac{5\pi}{6} $ is in the second quadrant, we have:

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

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

To find $ \tan(\frac{5\pi}{6}) $, use the identity $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $:

$$ \tan(\frac{5\pi}{6}) = \frac{\sin(\frac{5\pi}{6})}{\cos(\frac{5\pi}{6})} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$

Determine the angles that satisfy the tangent function

To determine the angles $\theta$ where $\tan(\theta) = \sqrt{3}$, we first recognize that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. From the unit circle, we know:

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

Thus, $\theta = \frac{\pi}{3}$ is one solution. Since tangent has a period of $\pi$, the general solution can be written as:

$$ \theta = \frac{\pi}{3} + n\pi $$

where $n$ is any integer.

Another solution within one period (from $0$ to $2\pi$) would be:

$$ \theta = \frac{4\pi}{3} $$

Therefore, the angles that satisfy $\tan(\theta) = \sqrt{3}$ within one period are $\theta = \frac{\pi}{3}$ and $\theta = \frac{4\pi}{3}$.

Find the Quadrant of a Point on the Unit Circle

To determine the quadrant of a point $(x, y)$ on the unit circle, we need to consider the signs of $x$ and $y$. The unit circle is centered at the origin (0,0) and has a radius of 1.

For example, let’s find which quadrant the point $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$ lies in:

1. Since $\frac{\sqrt{3}}{2} > 0$, the x-coordinate is positive.

2. Since $-\frac{1}{2} < 0$, the y-coordinate is negative.

From these observations, we know that the point lies in the fourth quadrant.

So, the point $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$ is in the fourth quadrant.

Fill in the missing value on the unit circle: What is the value of sin(θ) when cos(θ) = -1/2?

Given the unit circle equation:

$$\cos^2(θ) + \sin^2(θ) = 1$$

Substitute $\cos(θ) = -\frac{1}{2}$:

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

Simplify:

$$\frac{1}{4} + \sin^2(θ) = 1$$

Subtract $\frac{1}{4}$ from both sides:

$$\sin^2(θ) = 1 – \frac{1}{4}$$

$$\sin^2(θ) = \frac{3}{4}$$

Taking the square root of both sides:

$$\sin(θ) = \pm \sqrt{\frac{3}{4}}$$

$$\sin(θ) = \pm \frac{\sqrt{3}}{2}$$

So, the value of $\sin(θ)$ when $\cos(θ) = -\frac{1}{2}$ is $\pm \frac{\sqrt{3}}{2}$.