Math

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:

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

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

Calculate the cartesian coordinates of the intersection points of the unit circle and the line y = 2x + 1

First, the equation of the unit circle is:

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

Substitute $ y = 2x + 1 $ into $ x^2 + y^2 = 1 $:

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

Expand and simplify:

$$ x^2 + 4x^2 + 4x + 1 = 1 $$

Combine like terms:

$$ 5x^2 + 4x = 0 $$

Factor out:

$$ x(5x + 4) = 0 $$

Then, $ x = 0 $ or $ x = -\x0crac{4}{5} $.

For $ x = 0 $:

$$ y = 2(0) + 1 = 1 $$

So, one intersection point is $ (0, 1) $.

For $ x = -\x0crac{4}{5} $:

$$ y = 2(-\x0crac{4}{5}) + 1 = -\x0crac{8}{5} + 1 = -\x0crac{3}{5} $$

So, the other intersection point is $ (-\x0crac{4}{5}, -\x0crac{3}{5}) $.

The cartesian coordinates of the intersection points are $ (0, 1) $ and $ (-\x0crac{4}{5}, -\x0crac{3}{5}) $.

Find the exact value of sin(pi/4) on the unit circle

To find the exact value of $ \sin(\frac{\pi}{4}) $ on the unit circle, recognize that $ \frac{\pi}{4} $ is 45 degrees. The sine of 45 degrees (or $ \frac{\pi}{4} $) is:

$$ \sin(45^\circ) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} $$

Thus, the exact value is:

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

Find the exact value of tan(θ) on the unit circle for the angle θ where 0 < θ < π/2 and θ is the solution to the equation 2sin(θ)cos(θ) = 1

First, we start from the given equation:

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

We recognize that:

$$ 2\sin(\theta)\cos(\theta) = \sin(2\theta) $$

So the equation becomes:

$$ \sin(2\theta) = 1 $$

Since $ \sin(\frac{\pi}{2}) = 1 $, we have:

$$ 2\theta = \frac{\pi}{2} $$

Thus:

$$ \theta = \frac{\pi}{4} $$

Finally, we find that:

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

What is the sine of π/6?

The sine of $ \frac{π}{6} $ is found using the unit circle. At angle $ \frac{π}{6} $, the y-coordinate of the corresponding point on the unit circle is:

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

Evaluate the integral of cos(2x) divided by the square root of (1-sin^2(2x)) with respect to x

To evaluate the integral $ \int \frac{\cos(2x)}{\sqrt{1-\sin^2(2x)}} \, dx $, we begin by recognizing that:

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

Thus, the expression under the square root simplifies to:

$$ \sqrt{1-\sin^2(2x)} = \cos(2x) $$

Substituting this into the integral gives:

$$ \int \frac{\cos(2x)}{\cos(2x)} \, dx $$

This simplifies to:

$$ \int 1 \, dx $$

The integral of 1 with respect to $x$ is:

$$ x + C $$