Unit Circle

Given a point on the unit circle, find its cosine and sine values

Given a point \((\cos\theta, \sin\theta)\) on the unit circle, determine the coordinates when \(\theta = \frac{\pi}{4}\).

The unit circle has a radius of 1. At \(\theta = \frac{\pi}{4}\), both x and y coordinates are equal:

$$\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

Therefore, the coordinates are:

$$(\cos\frac{\pi}{4}, \sin\frac{\pi}{4}) = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$

Find the value of angle θ in degrees such that cos(θ) = sin(2θ) and θ lies in the interval [0, 360)

Given the equation:

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

We can use the double-angle identity for sine:

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

The equation becomes:

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

Dividing both sides by $\cos(\theta)$ (assuming $\cos(\theta) \neq 0$):

$$1 = 2\sin(\theta)$$

Solving for $\sin(\theta)$:

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

The values of $\theta$ in the interval [0, 360) where $\sin(\theta) = \frac{1}{2}$ are $\theta = 30^\circ$ and $\theta = 150^\circ$.

However, we also need to consider the case where $\cos(\theta) = 0$:

$\cos(\theta) = 0$ for $\theta = 90^\circ$ and $\theta = 270^\circ$.

Therefore, the angles that satisfy the equation are: $30^\circ$, $90^\circ$, $150^\circ$, and $270^\circ$.

Find the coordinates of the point on the unit circle at a given angle

To find the coordinates of the point on the unit circle at an angle $\theta$:

1. Use the parametric equations for the unit circle:

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

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

2. Substitute the given angle $\theta = \frac{2\pi}{3}$ into the equations:

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

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

Thus, the coordinates of the point are:

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

Find the point on the unit circle where the angle is π/3 and show all steps to verify the trigonometric coordinates

To find the point on the unit circle where the angle is $\frac{\pi}{3}$, we start by noting that the unit circle has a radius of 1. The coordinates of any point on the unit circle can be found using the trigonometric functions cosine (cos) and sine (sin).

We know that for an angle $\theta$:

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

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

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

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

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

Therefore, the coordinates of the point on the unit circle where the angle is $\frac{\pi}{3}$ are:

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

Find the coordinates of the point on the unit circle corresponding to the angle whose cosine is -2/3

Given the cosine of the angle is $-\frac{2}{3}$,

First, find the sine of the angle using the Pythagorean identity:

$$\cos^2\theta + \sin^2\theta = 1$$

Substitute $\cos\theta = -\frac{2}{3}$:

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

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

$$\sin^2\theta = 1 – \frac{4}{9}$$

$$\sin^2\theta = \frac{9}{9} – \frac{4}{9}$$

$$\sin^2\theta = \frac{5}{9}$$

Taking the square root,

$$\sin\theta = \pm\sqrt{\frac{5}{9}}$$

$$\sin\theta = \pm\frac{\sqrt{5}}{3}$$

Thus, the coordinates are:

$$(-\frac{2}{3}, \frac{\sqrt{5}}{3})$$ or $$(-\frac{2}{3}, -\frac{\sqrt{5}}{3})$$

Find the tangent of the angle where the unit circle intersects the x-axis at (1, 0)

To find the tangent of the angle, we first note that the point of intersection with the x-axis at (1, 0) corresponds to 0 radians or 0 degrees.

The tangent of an angle in a unit circle is given by $$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$.

For $$\theta = 0$$:

$$\sin(0) = 0$$ and $$\cos(0) = 1$$.

Therefore,

$$\tan(0) = \frac{0}{1} = 0$$.

So, the tangent of the angle is 0.

Find the sine and cosine of π/4 using the unit circle

To find the sine and cosine of $ \frac{\pi}{4} $ using the unit circle, we can use the coordinates of the corresponding point on the unit circle. For an angle of $ \frac{\pi}{4} $ radians, the coordinates are:

$ ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ) $

Therefore:

$$ \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

$$ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

What is the cosine of an angle in the unit circle corresponding to 7π/6 radians?

To find the cosine of the angle $ \frac{7\pi}{6} $ in the unit circle, we first recognize that this angle is in the third quadrant. An angle in the third quadrant has a negative cosine value.

The reference angle for $ \frac{7\pi}{6} $ is $ \frac{\pi}{6} $.

Since the cosine of $ \frac{\pi}{6} $ is $ \frac{\sqrt{3}}{2} $, the cosine of $ \frac{7\pi}{6} $ is $ -\frac{\sqrt{3}}{2} $.

Therefore, $ \cos(\frac{7\pi}{6}) = -\frac{\sqrt{3}}{2} $.

Find the coordinates of the point where the terminal side of an angle in standard position intersects the unit circle if the angle is given by theta = 7π/4

The unit circle has a radius of 1 and is centered at the origin. The coordinates (x, y) on the unit circle for an angle \( \theta \) are given by:

$$ (x, y) = (\cos(\theta), \sin(\theta)) $$

For \( \theta = \frac{7\pi}{4} \):

$$ \cos \left( \frac{7\pi}{4} \right) = \cos \left( 2\pi – \frac{\pi}{4} \right) = \cos \left( \frac{-\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

$$ \sin \left( \frac{7\pi}{4} \right) = \sin \left( 2\pi – \frac{\pi}{4} \right) = \sin \left( \frac{-\pi}{4} \right) = -\sin \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

Therefore, the coordinates are:

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

Given a point on the unit circle (a, b), find the value of cos(θ) and sin(θ)

Given a point on the unit circle $(a, b)$, we can find $\cos(\theta)$ and $\sin(\theta)$:

The coordinates of the point on the unit circle, $(a, b)$, represent the values of $\cos(\theta)$ and $\sin(\theta)$, respectively.

Thus,

$$ \cos(\theta) = a $$

$$ \sin(\theta) = b $$