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Find the coordinates of the point on the unit circle where the angle is 5π/4 radians

Find the coordinates of the point on the unit circle where the angle is 5π/4 radians

To find the coordinates of the point on the unit circle where the angle is $ \frac{5\pi}{4} $ radians, we can use the definitions of sine and cosine for the unit circle.

The angle $ \frac{5\pi}{4} $ is in the third quadrant, where both sine and cosine are negative.

For the unit circle, the coordinates are given by $(\cos \theta, \sin \theta)$.

Thus, we find:

$$ \cos \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

$$ \sin \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

Therefore, the coordinates are:

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

Find the coordinates of the point on the unit circle at an angle of 5π/4 radians

Find the coordinates of the point on the unit circle at an angle of 5π/4 radians

To find the coordinates of the point on the unit circle at an angle of $\frac{5\pi}{4}$ radians, we can use the cosine and sine functions:

$$ x = \cos \left( \frac{5\pi}{4} \right) $$

$$ y = \sin \left( \frac{5\pi}{4} \right) $$

First, let’s calculate the cosine value:

$$ \cos \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

Next, let’s calculate the sine value:

$$ \sin \left( \frac{5\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

Therefore, the coordinates of the point are:

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

Find the sine and cosine values for the angle 5π/6 on the unit circle, and determine the corresponding point on the unit circle

Find the sine and cosine values for the angle 5π/6 on the unit circle, and determine the corresponding point on the unit circle

First, we need to determine the reference angle for $\frac{5\pi}{6}$. The reference angle is $\pi – \frac{5\pi}{6} = \frac{\pi}{6}$.

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

The sine value for $\frac{\pi}{6}$ is $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

The cosine value for $\frac{\pi}{6}$ is $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

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

The corresponding point on the unit circle is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

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

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

Given an angle $\theta$, find the values of $\sin(\theta)$ and $\cos(\theta)$ using the unit circle equation. Suppose $\theta = \frac{\pi}{4}$.

The unit circle equation is given by:

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

For $\theta = \frac{\pi}{4}$, the corresponding point on the unit circle is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

Therefore,

$$\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \quad \text{and} \quad \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.$$

What is the cosine and sine of an angle of π/3 on the unit circle?

What is the cosine and sine of an angle of π/3 on the unit circle?

The angle $\frac{\pi}{3}$ corresponds to 60 degrees on the unit circle.

The coordinates of this angle on the unit circle are $(\frac{1}{2}, \frac{\sqrt{3}}{2})$.

Therefore, the cosine of the angle $\frac{\pi}{3}$ is $\frac{1}{2}$, and the sine is $\frac{\sqrt{3}}{2}$.

Find the coordinates on the unit circle for an angle of 5π/6

Find the coordinates on the unit circle for an angle of 5π/6

To find the coordinates of the point on the unit circle corresponding to an angle of $\frac{5\pi}{6}$, we can use the sine and cosine functions.

The angle $\frac{5\pi}{6}$ is in the second quadrant.

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

We know that for $\frac{\pi}{6}$:

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

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

Since $\frac{5\pi}{6}$ is in the second quadrant, the x-coordinate (cosine) will be negative and the y-coordinate (sine) will be positive.

Therefore, the coordinates are:

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

Find the sine and cosine of the angle θ = 45° using the unit circle

Find the sine and cosine of the angle θ = 45° using the unit circle

To find the sine and cosine of the angle $\theta = 45°$, we use the unit circle where the radius is 1.

The coordinates of the point where the terminal side of a 45° angle intersects the unit circle are $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

Therefore, $$\cos(45°) = \frac{\sqrt{2}}{2}$$ and $$\sin(45°) = \frac{\sqrt{2}}{2}$$.

Determine the Values of Trigonometric Functions at π/3

Determine the Values of Trigonometric Functions at π/3

Consider the angle $\frac{\pi}{3}$ on the unit circle. To find the values of sin, cos, and tan at this angle, we use the known values:

The sine of $\frac{\pi}{3}$ is:

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

The cosine of $\frac{\pi}{3}$ is:

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

Using the quotient identity for tangent:

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

Determine the cosine value of -π/3 using the unit circle

Determine the cosine value of -π/3 using the unit circle

First, recall that the unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. In the unit circle, the angle $\theta = -\pi/3$ is measured in the clockwise direction.

To find the cosine of $-\pi/3$, we can use the symmetry of the unit circle. The angle $-\pi/3$ is the same as $5\pi/3$ in the standard position (i.e., measured counterclockwise from the positive x-axis).

Cosine corresponds to the x-coordinate of the point on the unit circle. Thus, we need to find the x-coordinate of the point corresponding to $5\pi/3$.

At $5\pi/3$, the point on the unit circle is $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$. Therefore, the cosine of $-\pi/3$ is:

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

Find the value of tan(-π/6) using the unit circle

Find the value of tan(-π/6) using the unit circle

We start by recognizing that the angle $-\frac{\pi}{6}$ is equivalent to rotating $\frac{\pi}{6}$ radians in the clockwise direction.

On the unit circle, the point corresponding to $\frac{\pi}{6}$ radians is $(\frac{\sqrt{3}}{2}, \frac{1}{2})$. When we rotate in the clockwise direction to $-\frac{\pi}{6}$, the coordinates of the point become $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$.

The tangent of an angle is given by the ratio of the y-coordinate to the x-coordinate:

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

So, the value of $\tan(-\frac{\pi}{6})$ is $-\frac{\sqrt{3}}{3}$.

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