Homework

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

Evaluate the integral of cos^3(x)sin(x) with respect to x

To evaluate the integral of $ \cos^3(x)\sin(x) $ with respect to $ x $, we use a substitution method:

Let $ u = \cos(x) $, then $ du = -\sin(x) dx $. Consequently:

$$ \int \cos^3(x)\sin(x) dx = \int u^3 (-du) = -\int u^3 du $$

Now integrate:

$$ -\int u^3 du = -\frac{u^4}{4} + C $$

Substitute back $ \cos(x) $ for $ u $:

$$ -\frac{\cos^4(x)}{4} + C $$

The final answer is:

$$ -\frac{\cos^4(x)}{4} + C $$

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:

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

Determine the cosine of an angle given in radians and convert it to degrees

Given an angle $ \theta = \frac{7\pi}{6} $ radians, we need to determine its cosine and convert the angle to degrees.

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First, convert the angle to degrees:\n

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$$ \theta = \frac{7\pi}{6} \cdot \frac{180^\circ}{\pi} = 210^\circ $$

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The angle $ 210^\circ $ lies in the third quadrant where the cosine is negative.

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Using the unit circle, we know:

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$$ \cos(210^\circ) = \cos(180^\circ + 30^\circ) = -\cos(30^\circ) $$

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Since $ \cos(30^\circ) = \frac{\sqrt{3}}{2} $, we have:

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$$ \cos(210^\circ) = -\frac{\sqrt{3}}{2} $$

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Thus, the cosine of $ \frac{7\pi}{6} $ radians is $ -\frac{\sqrt{3}}{2} $ and the angle in degrees is $ 210^\circ $.

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 sine of an angle whose terminal side passes through the point (sqrt(3)/2, -1/2) on the unit circle

To find the sine of the angle, we need to identify the y-coordinate of the given point $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ on the unit circle.

The y-coordinate is:

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

Therefore, the sine of the angle is:

$$ \sin(\theta) = -\frac{1}{2} $$

Determine the quadrant for the given angle

The angle $ \theta = 45^\circ $ is in the first quadrant.