Homework

Determine the exact values of trigonometric functions at specific angles using the unit circle

To find the exact values of trigonometric functions for $ \theta = \frac{5\pi}{6} $, we first recognize that this angle corresponds to a reference angle of $ \frac{\pi}{6} $ in the second quadrant.

The coordinates of the point on the unit circle at $ \frac{\pi}{6} $ are $ (\frac{\sqrt{3}}{2}, \frac{1}{2}) $. Since $ \frac{5\pi}{6} $ lies in the second quadrant, the x-coordinate becomes negative:

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

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

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

Find the value of cos(θ) given the angle on the unit circle

Given that $\theta = \frac{5\pi}{6}$, find the value of $\cos(\theta)$ on the unit circle.

Step 1: Identify the reference angle.

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

Step 2: Determine the sign based on the quadrant.

$\theta = \frac{5\pi}{6}$ is in the second quadrant where cosine is negative.

Step 3: Find the value of cosine for the reference angle.

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

Step 4: Apply the sign from step 2.

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

How to remember the angles and coordinates on a Unit Circle

$$\text{To remember the angles and coordinates on a unit circle, follow these steps:}$$

$$1.\ \text{Divide the circle into four quadrants, each covering 90 degrees or } \frac{\pi}{2}$$

$$2.\ \text{Identify the key angles in radians: } 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \text{and} 2\pi$$

$$3.\ \text{Remember the coordinates for these key angles: } (1, 0), (\frac{\sqrt{3}}{2}, \frac{1}{2}), (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}), (\frac{1}{2}, \frac{\sqrt{3}}{2}), (0, 1), (-1, 0), (0, -1), \text{and back to} (1, 0)$$

$$4.\ \text{Use symmetry and reference angles to find the coordinates for other angles.}$$

Find the exact values of cosine and sine for the angle 7π/6 using the unit circle

To find the exact values of $\cos \frac{7\pi}{6}$ and $\sin \frac{7\pi}{6}$, we start by locating the angle on the unit circle. The angle $\frac{7\pi}{6}$ is in the third quadrant.

We know that $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$. This means the reference angle is $\frac{\pi}{6}$.

In the third quadrant, both the cosine and sine values are negative. The reference angle $\frac{\pi}{6}$ has known values of $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$.

Thus:

$$\cos \frac{7\pi}{6} = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$$

$$\sin \frac{7\pi}{6} = -\sin \frac{\pi}{6} = -\frac{1}{2}$$

Finding Sine, Cosine, and Tangent Values on the Unit Circle

Consider the angle $45^\circ$ (or $\frac{\pi}{4}$ radians) on the unit circle. Find the sine, cosine, and tangent values for this angle.

Step 1: Identify the coordinates on the unit circle for the angle $45^\circ$. The coordinates are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

Step 2: Using these coordinates, we can determine:

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

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

Step 3: Tangent is the ratio of sine to cosine:

$$\tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$$

Therefore,

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

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

$$\tan 45^\circ = 1$$

Find the values of cos(θ) for 3 different angles on the unit circle

To find the cosine values for angles on the unit circle, we first identify the angles and then use the unit circle definition.

Example angles: \(\theta = \frac{\pi}{3}, \theta = \frac{5\pi}{6}, \theta = \frac{7\pi}{4}\).

For \(\theta = \frac{\pi}{3}\):

Using the unit circle, we know that \(\cos(\frac{\pi}{3}) = \frac{1}{2}\).

For \(\theta = \frac{5\pi}{6}\):

Using the unit circle, we know that \(\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}\).

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

Using the unit circle, we know that \(\cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2}\).

Calculate the value of tan(θ) for θ = 7π/4 using the unit circle

To find the value of $\tan(\theta)$ for $\theta = \frac{7\pi}{4}$ using the unit circle, we first need to determine the coordinates of the point on the unit circle corresponding to $\theta = \frac{7\pi}{4}$.

$\theta = \frac{7\pi}{4}$ corresponds to an angle of $315^\circ$ in standard position.

In the unit circle, this point is $\left( \cos\left(\frac{7\pi}{4}\right), \sin\left(\frac{7\pi}{4}\right) \right)$.

The coordinates at $\frac{7\pi}{4}$ are $( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} )$.

Therefore, $\tan\left(\frac{7\pi}{4}\right) $ can be calculated as:

$$ \tan\left(\frac{7\pi}{4}\right) = \frac{\sin\left(\frac{7\pi}{4}\right)}{\cos\left(\frac{7\pi}{4}\right)} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$

Given a point on the unit circle, find its coordinates and the associated angle in radians, if the sine of the angle is equal to the cosine of the angle

Given $\sin(\theta) = \cos(\theta)$ for an angle $\theta$ on the unit circle:

We know that for an angle $\theta$ on the unit circle:

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

Let $\sin(\theta) = \cos(\theta) = x$. Then,

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

$$2x^2 = 1$$

$$x^2 = \frac{1}{2}$$

$$x = \pm \frac{1}{\sqrt{2}}$$

Therefore, $\sin(\theta) = \cos(\theta) = \pm \frac{1}{\sqrt{2}}$.

The coordinates are $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.

For $\frac{1}{\sqrt{2}}$, the angle is:

$$\theta = \frac{\pi}{4} + 2n\pi, \text{ for any integer } n$$

For $-\frac{1}{\sqrt{2}}$, the angle is:

$$\theta = \frac{5\pi}{4} + 2n\pi, \text{ for any integer } n$$

Find the coordinates of a point on the unit circle where the x-coordinate is 1/2

The equation of the unit circle is given by:

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

We are given that the x-coordinate is $\frac{1}{2}$. Substituting $x = \frac{1}{2}$ into the equation:

$$\left(\frac{1}{2}\right)^2 + y^2 = 1$$

$$\frac{1}{4} + y^2 = 1$$

Subtract $\frac{1}{4}$ from both sides:

$$y^2 = 1 – \frac{1}{4}$$

$$y^2 = \frac{3}{4}$$

Taking the square root of both sides:

$$y = \pm \sqrt{\frac{3}{4}}$$

$$y = \pm \frac{\sqrt{3}}{2}$$

Thus, the coordinates are:

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

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

To determine the coordinates of a point on the unit circle for a given angle $\theta$, we use the fact that the unit circle has a radius of 1 and the coordinates can be expressed as $(\cos(\theta), \sin(\theta))$.

Let’s find the coordinates for $\theta = \frac{\pi}{4}$.

The cosine and sine of $\frac{\pi}{4}$ are as follows:

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

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

Thus, the coordinates of the point are:

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