Homework

Find the coordinates of the point on the unit circle corresponding to an angle of \( \frac{5\pi}{4} \) radians

To find the coordinates of the point on the unit circle corresponding to the angle $ \frac{5\pi}{4} $ radians, we need to use the unit circle properties.

The angle $ \frac{5\pi}{4} $ radians is in the third quadrant.

The reference angle for $ \frac{5\pi}{4} $ is $ \pi/4 $ radians.

In the third quadrant, both sine and cosine values are negative.

From the unit circle, we know:

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

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

Therefore, the coordinates for $ \frac{5\pi}{4} $ are:

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

Calculate the cosine and sine of a 45-degree angle using the unit circle

To find the cosine and sine of a \(45^\circ\) angle, we use the unit circle, where the radius is 1.

In the unit circle, a \(45^\circ\) angle corresponds to \(\frac{\pi}{4}\) radians.

The coordinates of this point are \(\left( \cos \frac{\pi}{4}, \sin \frac{\pi}{4} \right)\).

For \(\frac{\pi}{4}\) radians:

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

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

Therefore, the cosine and sine of a 45-degree angle are both \(\frac{\sqrt{2}}{2}\).

Find the exact values of the cosine and sine for three points on the unit circle

Given points on the unit circle: $\frac{\pi}{6}, \frac{5\pi}{4}, \frac{11\pi}{6}$, find the exact values of $\cos$ and $\sin$.

Prove the relationship between the sine and cosine of the sum of two angles using the unit circle

To prove the relationship between the sine and cosine of the sum of two angles, we use the unit circle and the definitions of sine and cosine:

Given two angles, $\alpha$ and $\beta$, we can represent their sums on the unit circle. Consider the points $(\cos(\alpha), \sin(\alpha))$ and $(\cos(\beta), \sin(\beta))$.

Using the unit circle and the angle addition formulas, we have:

$$ \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) – \sin(\alpha) \sin(\beta) $$

$$ \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) $$

These relationships can be derived by examining the projections of the points on the unit circle and considering the definitions of sine and cosine in terms of coordinates.

Find the Coordinate on the Unit Circle

Given the angle $\theta = \frac{\pi}{4}$, find the coordinate on the unit circle.

The angle $\theta = \frac{\pi}{4}$ is equivalent to 45 degrees. At this angle, both the x and y coordinates are equal. Since we are on the unit circle, the coordinates can be determined by the values of $\cos\theta$ and $\sin\theta$.

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

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

Thus, the coordinate on the unit circle is:

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

Find the cosine and sine of an angle on the unit circle at \(\frac{5\pi}{6}\) radians

To find the cosine and sine values for $\frac{5\pi}{6}$ radians, first recognize that $\frac{5\pi}{6}$ is in the second quadrant of the unit circle.

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

Next, find the reference angle for $\frac{5\pi}{6}$, which is $\pi – \frac{5\pi}{6} = \frac{\pi}{6}$.

We know the sine and cosine values for the angle $\frac{\pi}{6}$:

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

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

Since $\frac{5\pi}{6}$ is in the second quadrant, the cosine value will be negative.

Therefore:

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

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

Given a point on the unit circle in Cartesian coordinates, find its other coordinate and the angle it makes with the positive x-axis

To solve this problem, we start by using the unit circle equation:

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

Given a point (x, y) = (\frac{1}{2}, y), we need to find y. Substitute x into the equation:

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

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

$$y^2 = 1 – \frac{1}{4}$$

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

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

So the points on the unit circle are (\frac{1}{2}, \frac{\sqrt{3}}{2}) and (\frac{1}{2}, -\frac{\sqrt{3}}{2}).

To find the angle with the positive x-axis:

$$\cos \theta = \frac{1}{2}$$

$$\theta = \pm \frac{\pi}{3}$$

Calculate the sine and cosine values for 225 degrees using the unit circle

The unit circle helps us determine the sine and cosine values for any given angle. For the angle $225^{\circ}$, we need to find its location on the unit circle.

The angle $225^{\circ}$ is in the third quadrant. In this quadrant, both sine and cosine values are negative. We can also express $225^{\circ}$ as $180^{\circ} + 45^{\circ}$, where $45^{\circ}$ is a reference angle.

From the unit circle, we know that the coordinates for $45^{\circ}$ are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$. Since $225^{\circ}$ is in the third quadrant, the sine and cosine values will be negative:

$$\cos(225^{\circ}) = -\frac{\sqrt{2}}{2}$$

$$\sin(225^{\circ}) = -\frac{\sqrt{2}}{2}$$

Thus, the sine and cosine values for $225^{\circ}$ are:

$$\cos(225^{\circ}) = -\frac{\sqrt{2}}{2}$$

$$\sin(225^{\circ}) = -\frac{\sqrt{2}}{2}$$

On the unit circle, find the value of cos(135°) + sin(225°) + tan(315°)

$$ \cos(135°) $$

Since $135°$ lies in the second quadrant, we have:

$$ \cos(135°) = -\cos(180° – 135°) = -\cos(45°) = -\frac{\sqrt{2}}{2} $$

$$ \sin(225°) $$

Since $225°$ lies in the third quadrant, we have:

$$ \sin(225°) = -\sin(360° – 225°) = -\sin(135°) = -\sin(180° – 135°) = -\sin(45°) = -\frac{\sqrt{2}}{2} $$

$$ \tan(315°) $$

Since $315°$ lies in the fourth quadrant, we have:

$$ \tan(315°) = \tan(360° – 45°) = \tan(45°) = 1 $$

Combining all these, we get:

$$ \cos(135°) + \sin(225°) + \tan(315°) = -\frac{\sqrt{2}}{2} + -\frac{\sqrt{2}}{2} + 1 = -\sqrt{2} + 1 $$

If the standard unit circle is flipped over the x-axis, describe the transformation of the angle θ and calculate the new coordinates for θ = 2π/3

When the unit circle is flipped over the x-axis, the y-coordinates of all points on the circle are inverted. Therefore, the angle $\theta$ remains the same in magnitude but the y-value of the coordinate changes sign.

For $\theta = \frac{2\pi}{3}$, the original coordinates on the unit circle are:

$$\left(\cos\left(\frac{2\pi}{3}\right), \sin\left(\frac{2\pi}{3}\right)\right)$$

We compute:

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

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

Therefore, the original coordinates are:
$$\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$

After flipping over the x-axis, the new coordinates become:
$$\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$$