Unit Circle

How to find the reference angle for any angle not on the unit circle

To find the reference angle for an angle θ not on the unit circle, you must first locate the angle in the appropriate quadrant. The reference angle is then the smallest angle between the terminal side of θ and the x-axis. Here are the steps:

1. If θ is in the first quadrant, the reference angle is θ itself:

$$ θ_{ref} = θ $$

2. If θ is in the second quadrant, the reference angle is:

$$ θ_{ref} = 180° – θ $$

3. If θ is in the third quadrant, the reference angle is:

$$ θ_{ref} = θ – 180° $$

4. If θ is in the fourth quadrant, the reference angle is:

$$ θ_{ref} = 360° – θ $$

Given that cos(θ) = -1/2, find the general solutions for θ in the unit circle

To solve for $ θ $ such that $ \cos(θ) = -\frac{1}{2} $, we need to find all angles in the unit circle where the cosine value is $ -\frac{1}{2} $. The cosine function is negative in the second and third quadrants.

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The reference angle for $ \cos(θ) = \frac{1}{2} $ is $ \frac{\pi}{3} $. Therefore, the general solutions in the second and third quadrants are:

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$$ θ = \pi – \frac{\pi}{3} + 2k\pi = \frac{2\pi}{3} + 2k\pi $$

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and

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$$ θ = \pi + \frac{\pi}{3} + 2k\pi = \frac{4\pi}{3} + 2k\pi $$

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where $ k $ is any integer.

How to memorize the coordinates of the unit circle

One method to memorize the unit circle is to remember key angles in radians and their corresponding coordinates. For example,:

$$ \text{At } \theta = 0, \ (1, 0) $$

$$ \text{At } \theta = \frac{\pi}{2}, \ (0, 1) $$

$$ \text{At } \theta = \pi, \ (-1, 0) $$

$$ \text{At } \theta = \frac{3\pi}{2}, \ (0, -1) $$

$$ \text{At } \theta = 2\pi, \ (1, 0) $$

These points divide the unit circle into four quadrants.

Prove that the integral of sin(x)/x from 0 to infinity is pi/2

To prove that the integral of $ \frac{\sin(x)}{x} $ from $ 0 $ to $ \infty $ is $ \frac{\pi}{2} $, we will use the fact that:

$$ \int_0^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} $$

The proof involves showing that the integral converges and evaluating it:

First, consider the function:

$$ f(t) = \int_0^t \frac{\sin(x)}{x} \, dx $$

As $ t \to \infty $, we must show that $ f(t) $ approaches $ \frac{\pi}{2} $. To do this, use the substitution $ x = t u $:

$$ \int_0^t \frac{\sin(x)}{x} \, dx = \int_0^1 \frac{\sin(tu)}{tu} \, t du = \int_0^1 \frac{\sin(tu)}{u} \, du $$

By integrating by parts and using properties of sine, we can show that:

$$ \int_0^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} $$

Solve for all x given that sin(x) + cos(2x) = 1, where x is an angle on the unit circle

To solve for all $x$ given that $\sin(x) + \cos(2x) = 1$:

First, use the double-angle identity for cosine: $\cos(2x) = 2\cos^2(x) – 1$

Substitute this into the equation:

$$ \sin(x) + 2\cos^2(x) – 1 = 1 $$

Rearrange the equation:

$$ \sin(x) + 2\cos^2(x) = 2 $$

Since $\sin^2(x) + \cos^2(x) = 1$, we have $\cos^2(x) = 1 – \sin^2(x)$

So:

$$ \sin(x) + 2(1 – \sin^2(x)) = 2 $$

Which simplifies to:

$$ \sin(x) + 2 – 2\sin^2(x) = 2 $$

Thus:

$$ \sin(x) – 2\sin^2(x) = 0 $$

Factor out $\sin(x)$:

$$ \sin(x)(1 – 2\sin(x)) = 0 $$

So $\sin(x) = 0$ or $\sin(x) = \frac{1}{2}$

Therefore, the solutions are:

$$ x = n\pi $$

and

$$ x = \frac{\pi}{6} + 2n\pi $$

or

$$ x = \frac{5\pi}{6} + 2n\pi $$

where $n$ is any integer.

Determine the exact value of a trigonometric expression involving radians on the unit circle

Consider the trigonometric expression $$ \cos\left(\frac{7\pi}{4}\right) + \sin\left(\frac{7\pi}{4}\right) $$. Determine its exact value using the unit circle.

First, convert the given angles to radians within the unit circle:

$$ \frac{7\pi}{4} $$ radians is equivalent to -$$ \frac{\pi}{4} $$ radians (since it is in the fourth quadrant).

The coordinates of the angle -$$ \frac{\pi}{4} $$ are given by:

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

Thus:

$$ \cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2} $$

$$ \sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} $$

Adding these values:

$$ \cos\left(\frac{7\pi}{4}\right) + \sin\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2} + \left(-\frac{\sqrt{2}}{2}\right) = 0 $$

Evaluate the integral of cos(x)sin(x) from 0 to pi/2

To evaluate the integral of $ \cos(x)\sin(x) $ from $ 0 $ to $ \frac{\pi}{2} $, we can use the substitution method. Let:

$$ u = \sin(x) $$

Then,

$$ du = \cos(x) dx $$

The integral transforms to:

$$ \int_0^{\frac{\pi}{2}} \cos(x)\sin(x) dx = \int_0^1 u du $$

Evaluating this integral:

$$ \int_0^1 u du = \frac{u^2}{2} \Bigg|_0^1 = \frac{1}{2} $$

Thus, the value of the integral is $ \frac{1}{2} $.

Explain how to derive the sine and cosine values of standard angles using the unit circle

To derive the sine and cosine values of standard angles (0°, 30°, 45°, 60°, and 90°) using the unit circle, follow these steps:

1. Draw the unit circle centered at the origin with a radius of 1.

2. Mark the standard angles on the unit circle. For example, angle 30° (or π/6) will be marked from the positive x-axis moving counter-clockwise.

3. For each angle, drop a perpendicular from the point on the unit circle to the x-axis. This forms a right triangle.

4. Use the definitions of sine and cosine: sine is the y-coordinate of the point, and cosine is the x-coordinate.

5. By using the properties of special triangles (such as 30°-60°-90° and 45°-45°-90° triangles), one can determine the exact coordinates of each point. For example, for 45° (or π/4), the coordinates are (√2/2, √2/2), so cos(45°) = √2/2 and sin(45°) = √2/2.

Evaluate the integral of the tangent function over the unit circle

To evaluate the integral of the function $ \tan(\theta) $ over the unit circle, we need to use the parametrization of the unit circle:

$$ x = \cos(\theta), \quad y = \sin(\theta), \quad d\theta $$

The integral over the unit circle in terms of $ \theta $ is:

$$ \int_{0}^{2\pi} \tan(\theta) \cdot \frac{dy}{d\theta} \ d\theta $$

Since $ \frac{dy}{d\theta} = \cos(\theta) $, we get:

$$ \int_{0}^{2\pi} \tan(\theta) \cos(\theta) \ d\theta $$

This integral can be simplified to:

$$ \int_{0}^{2\pi} \sin(\theta) \ d\theta = 0 $$

Find the sine and cosine values for $ \frac{5\pi}{4} $ on the unit circle

To find the sine and cosine values for $ \frac{5\pi}{4} $ on the unit circle, we need to locate the angle on the unit circle. The angle $ \frac{5\pi}{4} $ is in the third quadrant.

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

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

From the unit circle, we know that:

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

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

Thus, the sine and cosine values for $ \frac{5\pi}{4} $ are:

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

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