Unit Circle

Find the slope of the tangent line to the unit circle at the point where $\theta = \frac{\pi}{4}$

To find the slope of the tangent line to the unit circle at the point where $\theta = \frac{\pi}{4}$, we start by finding the coordinates of the point on the unit circle.

At $\theta = \frac{\pi}{4}$, the coordinates are:

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

The slope of the tangent line to the unit circle at any point $(x, y)$ is given by $-\frac{x}{y}$.

Therefore, the slope at the point $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ is:

$$ -\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$

Find the exact values of sin, cos, and tan for the angle 225° using the unit circle

To find the exact values of $\sin$, $\cos$, and $\tan$ for the angle $225^{\circ}$ using the unit circle, we first note that $225^{\circ}$ is in the third quadrant.

In the third quadrant, both sine and cosine values are negative, and tangent value is positive since tangent is the ratio of sine to cosine.

The reference angle for $225^{\circ}$ is $225^{\circ} – 180^{\circ} = 45^{\circ}$.

The values for $45^{\circ}$ are:

$$ \sin 45^{\circ} = \frac{\sqrt{2}}{2} $$

$$ \cos 45^{\circ} = \frac{\sqrt{2}}{2} $$

Therefore, the values in the third quadrant (for $225^{\circ}$) are:

$$ \sin 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \cos 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \tan 225^{\circ} = \frac{\sin 225^{\circ}}{\cos 225^{\circ}} = \frac{- \frac{\sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}} = 1 $$

Hence, the values are:

$$ \sin 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \cos 225^{\circ} = – \frac{\sqrt{2}}{2} $$

$$ \tan 225^{\circ} = 1 $$

Given that the point on the unit circle corresponding to the angle θ is (-3/5, -4/5), determine the value of θ in radians, ensuring that θ lies within the interval [0, 2π) Describe your method and calculations in detail

Given the point on the unit circle $\left( -\frac{3}{5}, -\frac{4}{5} \right)$, we need to determine the angle $\theta$ in radians.

First, note that the x and y coordinates tell us which quadrant the angle is in. Both coordinates are negative, so the point lies in the third quadrant.

The reference angle $\alpha$ can be determined using the tangent function:

$$ \tan \alpha = \left| \frac{y}{x} \right| = \left| \frac{-\frac{4}{5}}{-\frac{3}{5}} \right| = \frac{4}{3} $$

Using the arctangent function, we find:

$$ \alpha = \arctan \left( \frac{4}{3} \right) $$

Since this is a third quadrant angle, $\theta$ is given by:

$$ \theta = \pi + \alpha $$

Thus,

$$ \theta = \pi + \arctan \left( \frac{4}{3} \right) $$

Given a point P on the unit circle at an angle θ (in radians) from the positive x-axis, find the coordinates of P if θ is transformed by the function f(θ) = 2θ + π/4 Also, identify the new x and y coordinates after the transformation

Given the initial angle $ \theta $, the coordinates of the point $ P $ are:

$$ ( \cos \theta, \sin \theta ) $$

With the transformation $ f(\theta) = 2\theta + \frac{\pi}{4} $, let the new angle be $ \theta’ = 2\theta + \frac{\pi}{4} $. The new coordinates of the point $ P’ $ are:

$$ ( \cos(2\theta + \frac{\pi}{4}), \sin(2\theta + \frac{\pi}{4}) ) $$

For example, if $ \theta = \frac{\pi}{6} $, then:

$$ \theta’ = 2\times \frac{\pi}{6} + \frac{\pi}{4} = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12} $$

Thus, the new coordinates are:

$$ P’ ( \cos \frac{7\pi}{12}, \sin \frac{7\pi}{12} ) $$

Finding the Coordinates of a Point on the Unit Circle

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

First, recall the unit circle definition: for any angle $\theta$, the coordinates of the point on the unit circle are given by $(\cos \theta, \sin \theta)$. For $\theta = \frac{\pi}{3}$:

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

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

Thus, the coordinates are:

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

Find the sine and cosine values for the angle \(\theta = 45^{\circ}\) on the unit circle

To find the sine and cosine values for the angle $\theta = 45^{\circ}$ on the unit circle:

1. Note that $\theta = 45^{\circ}$ is in the first quadrant.

2. The coordinates of the corresponding point on the unit circle are given by $(\cos(45^{\circ}), \sin(45^{\circ}))$.

3. Using standard values:

$$\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$$

$$\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$$

Thus, the sine and cosine values for $\theta = 45^{\circ}$ are both $\frac{\sqrt{2}}{2}$.

Given that \( \csc(\theta) = 2 \) and \( \theta \) lies in the second quadrant, find the exact value of \( \theta \) and verify using trigonometric identities

Given:

$$ \csc(\theta) = 2 $$

Since \( \csc(\theta) = \frac{1}{\sin(\theta)} \), we get:

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

In the second quadrant, angle \( \theta \) where \( \sin(\theta) = \frac{1}{2} \) is:

$$ \theta = 180^\circ – 30^\circ = 150^\circ $$

Converting to radians:

$$ \theta = \pi – \frac{\pi}{6} = \frac{5\pi}{6} $$

Verification:

$$ \csc(\frac{5\pi}{6}) = \frac{1}{\sin(\frac{5\pi}{6})} = \frac{1}{\frac{1}{2}} = 2 $$

Thus, the exact value of \( \theta \) is:

$$ \boxed{\frac{5\pi}{6}} $$

What are the coordinates of the point on the unit circle corresponding to an angle of 45 degrees?

To determine the coordinates of the point on the unit circle at $45^\circ$, we use the fact that the unit circle has a radius of 1 and the coordinates are given by $ (\cos \theta, \sin \theta) $.

For $\theta = 45^\circ$, we have:

$$\cos 45^\circ = \frac{\sqrt{2}}{2}$$

$$\sin 45^\circ = \frac{\sqrt{2}}{2}$$

Therefore, the coordinates are:

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

Determine the coordinates of a point on a unit circle with a given angle

Let’s find the coordinates of a point on the unit circle corresponding to an angle of $\frac{5\pi}{4}$ radians.

The unit circle equation is given by:

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

For an angle $\theta$, the coordinates $(x, y)$ are given by:

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

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

Substituting $\theta = \frac{5\pi}{4}$:

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

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

Hence, the coordinates of the point are:

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

Find the value and angle for the given csc value

Given $csc(\theta) = \frac{5}{3}$, find the corresponding angle $\theta$.

We know:

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

Given,

$$\frac{1}{sin(\theta)} = \frac{5}{3}$$

So,

$$sin(\theta) = \frac{3}{5}$$

To find $\theta$, we take the inverse sine:

$$\theta = sin^{-1}(\frac{3}{5})$$

Using a calculator, we find:

$$\theta \approx 36.87^\circ \, or \, \theta \approx 143.13^\circ$$