Math

Determine the coordinates of the point on the unit circle corresponding to the angle 7π/6 radians

To find the coordinates of the point on the unit circle corresponding to the angle $\frac{7\pi}{6}$ radians, we need to consider the angle in standard position.

The angle $\frac{7\pi}{6}$ radians is in the third quadrant, where both sine and cosine are negative.

The reference angle is $\frac{\pi}{6}$ radians.

The coordinates for $\frac{\pi}{6}$ radians on the unit circle are $(\frac{\sqrt{3}}{2}, \frac{1}{2})$.

Since $\frac{7\pi}{6}$ is in the third quadrant, the coordinates are:

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

Find the cosine and sine of an angle on the unit circle

Given an angle $\theta = \frac{5\pi}{6}$ radians, find the coordinates $(\cos \theta, \sin \theta)$ on the unit circle.

Step 1: Recognize that $\theta = \frac{5\pi}{6}$ is an angle in the second quadrant.

Step 2: In the second quadrant, cosine is negative and sine is positive.

Step 3: Use the reference angle, which is $\pi – \frac{5\pi}{6} = \frac{\pi}{6}$.

Step 4: Recall the sine and cosine values for $\frac{\pi}{6}$: $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$.

Step 5: Apply the signs for the second quadrant: $\cos \frac{5\pi}{6} = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin \frac{5\pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2}$.

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

Find the value of cosecant for a given angle on the unit circle

Given that the angle \(\theta\) is \(\frac{5\pi}{6}\), find the value of \(csc(\theta)\) using the unit circle.

Step 1: First, locate the angle \(\frac{5\pi}{6}\) on the unit circle. This angle is in the second quadrant.

Step 2: The reference angle for \(\frac{5\pi}{6}\) is \(\frac{\pi}{6}\).

Step 3: The sine of \(\frac{5\pi}{6}\) is equal to the sine of \(\frac{\pi}{6}\) because they share the same reference angle.

Step 4: Thus, \(\sin(\frac{5\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}\).

Step 5: The cosecant function is the reciprocal of the sine function. Therefore, \(\csc(\frac{5\pi}{6}) = \frac{1}{\sin(\frac{5\pi}{6})} = \frac{1}{\frac{1}{2}} = 2\).

Find the cosecant of an angle in the unit circle

To find the $\csc(\theta)$ of an angle $\theta$ in the unit circle, first determine the sine of the angle. The cosecant is the reciprocal of the sine.

For example, consider $\theta = \frac{\pi}{6}$.

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

Therefore,

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

Given the point P(x, y) on the unit circle where x = -1/sqrt(2) and y = -1/sqrt(2), find the angle θ in radians

First, recognize that the coordinates given are $x = -\frac{1}{\sqrt{2}}$ and $y = -\frac{1}{\sqrt{2}}$. These values correspond to specific angles on the unit circle. We need to determine where both sine and cosine are negative and equal in magnitude.

Looking at the unit circle, we see that $\theta = \frac{5\pi}{4}$ radians has these properties.

Therefore, the angle $\theta$ in radians is \(\theta = \frac{5\pi}{4}\).

Find the cosine of the angle at the unit circle

To find the cosine of $$\theta = \frac{\pi}{3}$$ on the unit circle, we look at the x-coordinate of the point where the terminal side of the angle intersects the unit circle. For $$\theta = \frac{\pi}{3}$$, the point is $$ (\frac{1}{2}, \frac{\sqrt{3}}{2}) $$.

Therefore,

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

Find the coordinates of a point on the unit circle given an angle in radians

Given an angle $\theta = \frac{\pi}{3}$ radians, find the coordinates of the point on the unit circle.

The unit circle is defined by the equation $x^2 + y^2 = 1$. For an angle $\theta$, the coordinates $(x, y)$ can be found using:

$$x = \cos(\theta)$$

$$y = \sin(\theta)$$

Substituting $\theta = \frac{\pi}{3}$:

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

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

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

Find all angles θ in the interval [0, 2π) for which cotangent is equal to 1/√3 on the unit circle

To solve for $\theta$ in $\cot(\theta) = \frac{1}{\sqrt{3}}$, we start by recalling the definition of cotangent:

$$\cot(\theta) = \frac{1}{\tan(\theta)}$$

Given $\cot(\theta) = \frac{1}{\sqrt{3}}$, we have:

$$\frac{1}{\tan(\theta)} = \frac{1}{\sqrt{3}}$$

By taking reciprocals, we find:

$$\tan(\theta) = \sqrt{3}$$

The angles that satisfy $\tan(\theta) = \sqrt{3}$ on the unit circle are:

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

where $k$ is any integer. Considering the interval $[0, 2\pi)$, we get:

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

So, the angles are:

$$\boxed{\frac{\pi}{3}, \frac{4\pi}{3}}$$

Find the value of tan(θ) on the unit circle where θ=150°

To find $\tan(150^\circ)$, we first note that $150^\circ$ can be written as $180^\circ – 30^\circ$.

The reference angle here is $30^\circ$.

Since $\tan\theta$ is negative in the second quadrant:

$$\tan(150^\circ) = -\tan(30^\circ)$$

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}\ or \ \frac{\sqrt{3}}{3}$.

Therefore,

$$\tan(150^\circ) = -\frac{1}{\sqrt{3}}\ or \ -\frac{\sqrt{3}}{3}$$

Find the coordinates on the unit circle for an angle of \(\frac{\pi}{3}\) radians

To find the coordinates of an angle of $\frac{\pi}{3}$ radians on the unit circle, we use the cosine and sine values of the angle.

The cosine of $\frac{\pi}{3}$ is $\frac{1}{2}$.

The sine of $\frac{\pi}{3}$ is $\frac{\sqrt{3}}{2}$.

Therefore, the coordinates are:

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