Homework

Find the secant line to the unit circle that is equidistant from the x-axis

To find the secant line to the unit circle that is equidistant from the $x$-axis, we use the equation of the unit circle

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

and the general equation of a line

$$ y = mx + b $$

Since the secant line is equidistant from the $x$-axis, the $y$-intercept $b$ must satisfy the condition that the distances from $b$ to the points of intersection with the circle are equal. So, we solve:

Substitute $y = mx + b$ into the circle

Determine the exact values of sine and cosine for the angle π/4 using the unit circle

To find the exact values of sine and cosine for the angle $ \frac{\pi}{4} $, we use the unit circle.

For $ \theta = \frac{\pi}{4} $, the coordinates on the unit circle are:

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

Since $ \frac{\pi}{4} $ is an angle in the first quadrant where sine and cosine values are positive, we use the 45-degree reference angle values. We have:

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

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

Thus, the exact values are:

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

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

Identify the coordinates of specific angles on the unit circle

To find the coordinates of specific angles on the unit circle, remember that the unit circle has a radius of 1.

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

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

Find the area of a sector in a unit circle with a central angle of θ radians

The formula to find the area of a sector in a unit circle is:

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

where $ \theta $ is the central angle in radians.

For example, if $ \theta = \frac{\pi}{4} $:

$$ A = \frac{1}{2} \times \frac{\pi}{4} = \frac{\pi}{8} $$

Find the image of the unit circle under the transformation f(z)=z^2

The unit circle in the complex plane is given by $ |z| = 1 $, meaning any point $ z $ on the unit circle can be written as $ z = e^{i\theta} $ for some real number $ \theta $.

Under the transformation $ f(z) = z^2 $, the image of $ z $ is:

$$ f(z) = (e^{i\theta})^2 = e^{i(2\theta)} $$

Since $ e^{i(2\theta)} $ is still a point on the unit circle, the image of the unit circle under $ f(z) = z^2 $ is the unit circle itself.

Identify the quadrant in which the angle lies

To identify the quadrant in which the angle $ \theta $ lies, follow these steps:

1. If $ 0 \leq \theta < \frac{\pi}{2} $, then the angle is in the first quadrant.

2. If $ \frac{\pi}{2} \leq \theta < \pi $, then the angle is in the second quadrant.

3. If $ \pi \leq \theta < \frac{3\pi}{2} $, then the angle is in the third quadrant.

4. If $ \frac{3\pi}{2} \leq \theta < 2\pi $, then the angle is in the fourth quadrant.

Find the value of cos(π/3)

The value of $ \cos\left( \frac{\pi}{3} \right) $ can be found using the unit circle. The angle $ \frac{\pi}{3} $ corresponds to 60 degrees. On the unit circle, the coordinates for the angle 60 degrees are:

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

Therefore, the value of $ \cos\left( \frac{\pi}{3} \right) $ is:

$$ \frac{1}{2} $$

Find the coordinates of a point on the unit circle at angle π/4

To find the coordinates of a point on the unit circle at angle $\frac{\pi}{4}$, we use the trigonometric functions:

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

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

Since:

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

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

The coordinates are:

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

Find the values of sine, cosine, and tangent for the angle 5π/6 using the unit circle

To find the values of $ \sin \frac{5\pi}{6} $, $ \cos \frac{5\pi}{6} $, and $ \tan \frac{5\pi}{6} $ using the unit circle, we first locate the angle $ \frac{5\pi}{6} $ on the unit circle.

The angle $ \frac{5\pi}{6} $ is in the second quadrant, where sine is positive and cosine is negative. The reference angle is $ \pi – \frac{5\pi}{6} = \frac{\pi}{6} $.

From the unit circle, we know:

$$ \sin \frac{\pi}{6} = \frac{1}{2} $$ and $$ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $$

Since we are in the second quadrant:

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

$$ \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} $$

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

Find the points where the line y = mx + b is tangent to the unit circle

To find the points where the line $ y = mx + b $ is tangent to the unit circle, we start with the equation of the unit circle:

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

Substitute $ y = mx + b $ into the unit circle equation:

$$ x^2 + (mx + b)^2 = 1 $$

Expand the equation:

$$ x^2 + m^2x^2 + 2mxb + b^2 = 1 $$

Combine like terms:

$$ (1 + m^2)x^2 + 2mxb + b^2 = 1 $$

This is a quadratic equation in $ x $:

$$ (1 + m^2)x^2 + 2mxb + (b^2 – 1) = 0 $$

For the line to be tangent to the circle, the discriminant must be zero:

$$ (2mb)^2 – 4(1 + m^2)(b^2 – 1) = 0 $$

Simplify the discriminant:

$$ 4m^2b^2 – 4(1 + m^2)(b^2 – 1) = 0 $$

Solving this will give the condition on $ b $.