Math

Find the sine and cosine values for the angle θ = 30° on the unit circle

To find the sine and cosine values for $ \theta = 30° $ on the unit circle, we need to locate the point on the unit circle corresponding to $ \theta = 30° $.

1. Convert degrees to radians, $ \theta = 30° = \frac{π}{6} $ radians.

2. From the unit circle, the coordinates for $ \theta = \frac{π}{6} $ are $ ( \cos(30°), \sin(30°) ) $.

3. Hence, $ \cos(30°) = \frac{\sqrt{3}}{2} $ and $ \sin(30°) = \frac{1}{2} $.

So, the cosine value is $ \frac{\sqrt{3}}{2} $ and the sine value is $ \frac{1}{2} $.

Find the value of tan(θ) for θ in the unit circle

To find the value of $\tan(\theta)$ for $\theta$ on the unit circle, consider the point $P(x, y)$ on the circle corresponding to the angle $\theta$. The tangent of $\theta$ is given by the ratio of the y-coordinate to the x-coordinate, i.e., $\tan(\theta) = \frac{y}{x}$. For example, if $\theta = \frac{\pi}{4}$, the coordinates of the corresponding point on the unit circle are $\left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$. Therefore,

$$\tan \left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1.$$

Determine the Quadrant of a Given Point on a Unit Circle

Given the point \((x, y)\) on a unit circle, determine the quadrant in which the point lies.

The unit circle has a radius of 1. The quadrants are defined as follows:

– Quadrant I: \((x > 0, y > 0)\)

– Quadrant II: \((x < 0, y > 0)\)

– Quadrant III: \((x < 0, y < 0)\)

– Quadrant IV: \((x > 0, y < 0)\)

Let’s solve for the point \((-\frac{1}{2}, \frac{\sqrt{3}}{2})\)

Given: \(x = -\frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\)

Since \(x < 0\) and \(y > 0\), the point lies in Quadrant II.

Find the value of tan(θ) at three specific angles on the unit circle: θ = π/4, 3π/4, and 5π/6

For $\theta = \frac{\pi}{4}$:

On the unit circle, at $\theta = \frac{\pi}{4}$, the coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

The tangent function is given by $\tan(\theta) = \frac{y}{x}$.

Thus,
$$ \tan(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

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

On the unit circle, at $\theta = \frac{3\pi}{4}$, the coordinates are $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

Thus,
$$ \tan(\frac{3\pi}{4}) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

For $\theta = \frac{5\pi}{6}$:

On the unit circle, at $\theta = \frac{5\pi}{6}$, the coordinates are $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$.

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

If $\theta$ is an angle in the unit circle such that $\cos(\theta) = \frac{1}{2}$ and $\sin(\theta) = \frac{\sqrt{3}}{2}$, find the value of $\theta$ in degrees and radians

To solve this problem, we need to determine the angle $\theta$ on the unit circle. Given:

$$\cos(\theta) = \frac{1}{2}$$ $$\sin(\theta) = \frac{\sqrt{3}}{2}$$

On the unit circle, these values correspond to the angle $\theta = 60^{\circ}$ or $\theta = \frac{\pi}{3}$ radians.

Therefore, the value of $\theta$ is:

$$\theta = 60^{\circ}$$ $$\theta = \frac{\pi}{3}$$

Find the exact value of cos(5π/6) using the unit circle

To find the exact value of $\cos(\frac{5\pi}{6})$, we first determine the location of the angle on the unit circle.

The angle $\frac{5\pi}{6}$ is in the second quadrant. In the unit circle, the cosine of an angle in the second quadrant is negative.

The reference angle for $\frac{5\pi}{6}$ is $\pi – \frac{5\pi}{6}$, which simplifies to $\frac{\pi}{6}$.

The cosine of $\frac{\pi}{6}$ is $\frac{\sqrt{3}}{2}$. Therefore, $\cos(\frac{5\pi}{6}) = – \frac{\sqrt{3}}{2}$.

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

Find the angle on the unit circle corresponding to the coordinates (-2/3, y)

To find the angle on the unit circle corresponding to the coordinates $\left(-\frac{2}{3}, y\right)$, we need to use the Pythagorean identity:

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

Since $x = -\frac{2}{3}$, we plug this value into the equation:

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

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

Subtract $\frac{4}{9}$ from both sides:

$$y^2 = 1 – \frac{4}{9}$$

$$y^2 = \frac{9}{9} – \frac{4}{9}$$

$$y^2 = \frac{5}{9}$$

Take the square root of both sides:

$$y = \pm\sqrt{\frac{5}{9}}$$

$$y = \pm\frac{\sqrt{5}}{3}$$

The coordinates are $\left(-\frac{2}{3}, \pm\frac{\sqrt{5}}{3}\right)$.

Given the unit circle, find the angles θ between 0 and 2π for which the secant function sec(θ) equals 2, and provide a step-by-step explanation for your answer

We start by recalling the definition of the secant function: $$\sec(\theta) = \frac{1}{\cos(\theta)}$$. Therefore, the given condition $\sec(\theta) = 2$ translates to:

$$\frac{1}{\cos(\theta)} = 2$$

Solving for $\cos(\theta)$, we get:

$$\cos(\theta) = \frac{1}{2}$$

Now, we need to find the angles $\theta$ in the interval $[0, 2\pi)$ such that $\cos(\theta) = \frac{1}{2}$. These angles can be found using the unit circle:

$$\theta = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2k\pi \quad \text{for integers k}$$

Considering the interval $0 \leq \theta < 2\pi$, we have:

$$\theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{5\pi}{3}$$

Therefore, the angles $\theta$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

Find the coordinates of points where the angle is 2π/3 on the unit circle

To find the coordinates of the points where the angle is $$ \frac{2\pi}{3} $$ on the unit circle, we use the unit circle definition where any point can be given by $(\cos(\theta), \sin(\theta))$.

Here, $$ \theta = \frac{2\pi}{3} $$.

Therefore, the coordinates are:

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

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

Thus, the coordinates are:

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

Determine the coordinates of the point on the unit circle for an angle of 5π/6 radians Also, find the corresponding angle in degrees

To determine the coordinates of the point on the unit circle corresponding to an angle of $\frac{5\pi}{6}$ radians, we follow these steps:

1. Convert the angle into degrees:

$$\frac{5\pi}{6} \times \frac{180}{\pi} = 150^\circ$$

2. Find the coordinates using trigonometric functions on the unit circle:

$$x = \cos(150^\circ) = \cos(180^\circ – 30^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}$$

$$y = \sin(150^\circ) = \sin(180^\circ – 30^\circ) = \sin(30^\circ) = \frac{1}{2}$$

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

The corresponding angle in degrees is $$150^\circ$$.