Homework

Find the values of tan(θ), sin(θ), and cos(θ) for θ = 45 degrees

To find the values of $\tan(\theta)$, $\sin(\theta)$, and $\cos(\theta)$ for $\theta = 45^\circ$:

First, we note that $\theta = 45^\circ$ is in the first quadrant of the unit circle.

The coordinates of the point on the unit circle at $\theta = 45^\circ$ are:

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

Therefore:

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

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

Using the definition of tangent:

$$\tan(45^\circ) = \frac{\sin(45^\circ)}{\cos(45^\circ)} = 1$$

Prove the identity involving cos and sin on the unit circle

To prove the identity involving $ \cos(\theta) $ and $ \sin(\theta) $ on the unit circle, we start with the Pythagorean identity:

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

Consider the parameterization of the unit circle with $ \theta $ as the angle from the positive x-axis:

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

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

Then, the coordinates $ (x, y) $ must satisfy:

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

Substituting $ x = \cos(\theta) $ and $ y = \sin(\theta) $, we get:

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

This verifies the identity.

Find the coordinates of the point where the line y = 1 intersects the unit circle

To find the coordinates where the line $ y = 1 $ intersects the unit circle, we start by recalling the equation of the unit circle:

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

Substituting $ y = 1 $ into the unit circle equation, we get:

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

Simplifying,

$$ x^2 + 1 = 1 $$

$$ x^2 = 0 $$

$$ x = 0 $$

Therefore, the point of intersection is:

$$ (0, 1) $$

Find the exact values of sin, cos, and tan at 30 degrees on the unit circle

First, we need to convert $ 30^{\circ} $ to radians:

$$ 30^{\circ} = \frac{\pi}{6} $$

Using the unit circle, the coordinates for $ \frac{\pi}{6} $ are $ \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) $

From this, we can find:

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

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

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

Determine the value of tan(θ) when sin(θ) = 3/5 and θ is in the first quadrant

Given that $\sin(θ) = \frac{3}{5}$ and $θ$ is in the first quadrant, we can find $\cos(θ)$ using the Pythagorean identity:

$$\sin^2(θ) + \cos^2(θ) = 1$$

Plugging in the given value:

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

$$\frac{9}{25} + \cos^2(θ) = 1$$

$$\cos^2(θ) = 1 – \frac{9}{25} = \frac{16}{25}$$

Since $θ$ is in the first quadrant, $\cos(θ)$ is positive:

$$\cos(θ) = \frac{4}{5}$$

Now, we can find $\tan(θ)$:

$$\tan(θ) = \frac{\sin(θ)}{\cos(θ)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$$

Therefore, $\tan(θ) = \frac{3}{4}$.

Find the values of tan(θ) at various angles and verify using the unit circle

To find the values of $ \tan(\theta) $ at various angles and verify using the unit circle, we consider the following angles: $ \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} $

1. For $ \theta = \frac{\pi}{4} $:

$$ \tan(\frac{\pi}{4}) = 1 $$

2. For $ \theta = \frac{3\pi}{4} $:

$$ \tan(\frac{3\pi}{4}) = -1 $$

3. For $ \theta = \frac{5\pi}{4} $:

$$ \tan(\frac{5\pi}{4}) = 1 $$

4. For $ \theta = \frac{7\pi}{4} $:

$$ \tan(\frac{7\pi}{4}) = -1 $$

Verification: Using the unit circle, we observe that at these angles, the tangent value is consistent with the coordinates (x, y) where $ \tan(\theta) = \frac{y}{x} $.

Determine the coordinates of a point on the unit circle where the tangent line has a slope of 3/4

To determine the coordinates of a point on the unit circle where the tangent line has a slope of $\frac{3}{4}$, we start with the equation of the unit circle:

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

The slope of the tangent line at a point $(x, y)$ on the circle can be found by differentiating implicitly:

$$2x + 2y\frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$, we get:

$$\frac{dy}{dx} = -\frac{x}{y}$$

We need the slope to equal $\frac{3}{4}$:

$$-\frac{x}{y} = \frac{3}{4}$$

This implies:

$$y = -\frac{4x}{3}$$

Substitute $y = -\frac{4x}{3}$ back into the unit circle equation:

$$x^2 + \left(-\frac{4x}{3}\right)^2 = 1$$

$$x^2 + \frac{16x^2}{9} = 1$$

$$\frac{25x^2}{9} = 1$$

$$x^2 = \frac{9}{25}$$

$$x = \pm \frac{3}{5}$$

Substitute $x$ back into $y = -\frac{4x}{3}$:

$$y = \mp \frac{4 \cdot \frac{3}{5}}{3} = \mp \frac{4}{5}$$

The coordinates are:

$$\left(\frac{3}{5}, -\frac{4}{5}\right)$$ and $$\left(-\frac{3}{5}, \frac{4}{5}\right)$$

Determine the angle in radians of the point on the unit circle in the first quadrant with an x-coordinate of 1/2

To find the angle in radians with an $x$-coordinate of $\frac{1}{2}$ in the first quadrant, we use the unit circle definition of cosine.

For $\cos(\theta) = \frac{1}{2}$, the corresponding angle is:

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

Find the value of sec(θ) for θ in the unit circle

To find the value of $ \sec(\theta) $ for $ \theta $ in the unit circle, we need to recall the definition of secant. The secant function is the reciprocal of the cosine function:

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

Given that $ \theta $ is an angle in the unit circle, let’s consider $ \theta = \frac{\pi}{4} $ as an example. For this angle:

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

Thus,

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

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