Homework

What does sin represent on the unit circle?

On the unit circle, the sine of an angle represents the y-coordinate of the point on the unit circle that corresponds to that angle.

$$\text{Given an angle } \theta, \text{the coordinates of the point P on the unit circle are } (\cos(\theta), \sin(\theta)).$$

This means:

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

For example, if \(\theta = 30^\circ\):

$$\sin(30^\circ) = \frac{1}{2}.$$

Find the value of sec(θ) if point P(1/2, √3/2) lies on the unit circle

To find $\sec(\theta)$, we need to know $\cos(\theta)$. Given the coordinates on the unit circle, $\cos(\theta) = x$-coordinate of point $P$.

Here, $x = \frac{1}{2}$. Therefore, $\cos(\theta) = \frac{1}{2}$.

Recall that $\sec(\theta) = \frac{1}{\cos(\theta)}$.

Thus, $\sec(\theta) = \frac{1}{\frac{1}{2}} = 2$.

Therefore, $\sec(\theta) = 2$.

Find the coordinates of the point where the terminal side of a 225-degree angle intersects the unit circle

Given an angle of $225^{\circ}$, we first convert it to radians:

$$225^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{5\pi}{4} \text{ radians}$$

The angle $\frac{5\pi}{4}$ is in the third quadrant, where both sine and cosine are negative. The reference angle is $225^{\circ} – 180^{\circ} = 45^{\circ}$. Since $\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$ and $\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$, we have:

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

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

Thus, the coordinates are:

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

What Does Sin Represent on the Unit Circle?

On the unit circle, the function $\sin$ represents the y-coordinate of a point on the circle. The unit circle is defined as a circle with a radius of 1, centered at the origin of the coordinate system.

First, consider a point on the unit circle defined by an angle $\theta$, measured in radians from the positive x-axis. This point can be represented as $(\cos(\theta), \sin(\theta))$.

Because the radius of the unit circle is 1, the coordinates $(\cos(\theta), \sin(\theta))$ correspond to the horizontal and vertical distances from the origin.

Therefore,

$$\sin(\theta)$$

represents the vertical distance from the x-axis to the point on the unit circle. For example, if $\theta = \frac{\pi}{6}$ (30 degrees), the point on the unit circle is $(\frac{\sqrt{3}}{2}, \frac{1}{2})$. Thus, $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Find the coordinates of the points where the line y = x intersects the unit circle

We start with the unit circle equation:

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

Substituting $y = x$, we get:

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

$$2x^2 = 1$$

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

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

Since $y = x$, the coordinates are:

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

and

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

Determine the values of sin(θ) and cos(θ) for θ = 5π/6

Let $θ = \frac{5π}{6}$. This angle is in the second quadrant.

To find $\sin(θ)$ and $\cos(θ)$, we use the reference angle $θ’ = π – \frac{5π}{6} = \frac{π}{6}$.

The sine and cosine of $\frac{π}{6}$ are:

$$\sin\left(\frac{π}{6}\right) = \frac{1}{2}, \cos\left(\frac{π}{6}\right) = \frac{\sqrt{3}}{2}$$

Since the angle is in the second quadrant, $\sin(θ)$ is positive and $\cos(θ)$ is negative.

Thus,

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

On the unit circle, what is the value of sin(π/6)?

To find the value of $\sin(\frac{\pi}{6})$ on the unit circle, we start by understanding that $\frac{\pi}{6}$ radians is equivalent to 30 degrees. In a unit circle, the coordinates of the angle $\frac{\pi}{6}$ are (\(\frac{\sqrt{3}}{2}, \frac{1}{2}\)).

Therefore, $\sin(\frac{\pi}{6})$ is the y-coordinate, which is:

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

Find the value of tan(θ) given that sin(θ) = 3/5 and θ is in the second quadrant Show all steps

Given that $\sin(\theta) = \frac{3}{5}$, we know the following:

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

$$\left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1$$

$$\frac{9}{25} + \cos^2(\theta) = 1$$

$$\cos^2(\theta) = 1 – \frac{9}{25}$$

$$\cos^2(\theta) = \frac{16}{25}$$

Since $\theta$ is in the second quadrant, $\cos(\theta)$ is negative:

$$\cos(\theta) = -\frac{4}{5}$$

Now, $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$

$$\tan(\theta) = \frac{\frac{3}{5}}{-\frac{4}{5}}$$

$$\tan(\theta) = -\frac{3}{4}$$

So, $\tan(\theta) = -\frac{3}{4}$.

Find the exact values of sin(θ) and cos(θ) for θ = 3π/4 using the unit circle

To find the exact values of $\sin(\theta)$ and $\cos(\theta)$ for $\theta = \frac{3\pi}{4}$, we can use the unit circle.

First, note that $\theta = \frac{3\pi}{4}$ is in the second quadrant. In the unit circle, the angle $\frac{3\pi}{4}$ corresponds to $135^\circ$.

For angles in the second quadrant, the sine value is positive and the cosine value is negative. The reference angle for $\frac{3\pi}{4}$ is $\frac{\pi}{4}$ (or $45^\circ$), where $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

Therefore, $\sin(\frac{3\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{3\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$.

So, the exact values are $\sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$.