Unit Circle

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:

$$ x + C $$

Find the secant of an angle θ in a unit circle

To find the secant of an angle $\theta$ in a unit circle, we use the formula:

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

Suppose $\theta$ is an angle in the first quadrant where cos(θ) = 0.6. Then:

$$ \sec(\theta) = \frac{1}{0.6} = \frac{5}{3} $$

Calculate the tangent of an angle when given the sine and cosine values in the unit circle

To find the tangent of an angle in the unit circle when given the sine and cosine values, we use the formula:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$

For example, if $\sin(\theta) = \frac{1}{2}$ and $\cos(\theta) = \frac{\sqrt{3}}{2}$, then:

$$ \tan(\theta) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

Determine the values of cos(theta) and sin(theta) given that the point (x, y) is on the unit circle

Given that $ (x, y) $ is on the unit circle, we know:

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

Using the definitions of the trigonometric functions on the unit circle, we have:

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

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

Thus, the values of $ \cos(\theta) $ and $ \sin(\theta) $ are:

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

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

Define a unit circle and prove that any point (x, y) on the unit circle satisfies the equation x^2 + y^2 = 1

A unit circle is a circle with a radius of 1 unit, centered at the origin (0,0) in the Cartesian coordinate system.

To prove that any point $ (x, y) $ on the unit circle satisfies $ x^2 + y^2 = 1 $, we start with the definition of a circle:

$$ (x – h)^2 + (y – k)^2 = r^2 $$

For a unit circle, the center is at (0, 0) and the radius $ r $ is 1, so the equation becomes:

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

Thus, any point $ (x, y) $ on the unit circle will satisfy this equation.

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

To find the exact value of $ \sin(\frac{\pi}{4}) $ on the unit circle, we recognize that $ \frac{\pi}{4} $ is equivalent to $ 45^{\circ} $.

On the unit circle, the coordinates for $ \frac{\pi}{4} $ are $ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $.

The sine value is the y-coordinate, so:

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