Math

Find the exact values of sin(3π/4), cos(3π/4), and tan(3π/4) using the unit circle

We are asked to find the exact values of $\sin(\frac{3\pi}{4})$, $\cos(\frac{3\pi}{4})$, and $\tan(\frac{3\pi}{4})$ using the unit circle.

First, we locate the angle $\frac{3\pi}{4}$ on the unit circle: it is in the second quadrant.

The reference angle for $\frac{3\pi}{4}$ is $\frac{\pi}{4}$ (45 degrees). In the second quadrant, the sine value is positive, and the cosine value is negative.

Thus, we have:

$$\sin(\frac{3\pi}{4}) = \sin(\pi – \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$

$$\cos(\frac{3\pi}{4}) = \cos(\pi – \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$

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

Find the angle on the unit circle in the complex plane where the cosine value is 1/2

We start by knowing that the cosine function gives the real part of the point on the unit circle corresponding to a given angle.

We are given $\cos(\theta) = \frac{1}{2}$ and need to find the angles $\theta$ where this holds true.

On the unit circle, $\cos(\theta)$ reaches $\frac{1}{2}$ at two points: $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$.

Therefore, the angles are:

$$\theta = \frac{\pi}{3}, \frac{5\pi}{3}$$

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

Convert the angle 150 degrees to radians and find its coordinates on the unit circle

To convert degrees to radians, we use the conversion factor \( \frac{\pi}{180} \).

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

The coordinates on the unit circle for \( \theta = \frac{5\pi}{6} \) are given by \((\cos(\theta), \sin(\theta))\).

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

Thus, the coordinates are:

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

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 sine and cosine values at different points on the unit circle

To find the sine and cosine values at different points on the unit circle, we can use the angle in radians.

1. For the angle $$\frac{\pi}{6}$$ radians:

The coordinates on the unit circle are $$\left( \cos \frac{\pi}{6}, \sin \frac{\pi}{6} \right)$$.

Thus, we get: $$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$$ and $$\sin \frac{\pi}{6} = \frac{1}{2}$$.

2. For the angle $$\frac{\pi}{4}$$ radians:

The coordinates on the unit circle are $$\left( \cos \frac{\pi}{4}, \sin \frac{\pi}{4} \right)$$.

Thus, we get: $$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$ and $$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$.

3. For the angle $$\frac{\pi}{3}$$ radians:

The coordinates on the unit circle are $$\left( \cos \frac{\pi}{3}, \sin \frac{\pi}{3} \right)$$.

Thus, we get: $$\cos \frac{\pi}{3} = \frac{1}{2}$$ and $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$.

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

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