Unit Circle

Find the sine and cosine of the angle θ on the unit circle when θ = 5π/4

To find the sine and cosine of the angle $\theta = \frac{5\pi}{4}$ on the unit circle, we use the definitions of the trigonometric functions on the unit circle. The angle $\frac{5\pi}{4}$ is in the third quadrant.

For angles in the third quadrant, both sine and cosine are negative. The reference angle for $\theta = \frac{5\pi}{4}$ is $\frac{\pi}{4}$.

The sine and cosine of $\frac{\pi}{4}$ are both $\frac{\sqrt{2}}{2}$.

Thus:

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

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

Find the coordinates of a point on the unit circle that corresponds to an angle of π/3 radians

To find the coordinates of a point on the unit circle corresponding to an angle of $ \frac{\pi}{3} $ radians:

We can use the unit circle definitions for sine and cosine.

$$ x = \cos \left( \frac{\pi}{3} \right) $$

$$ y = \sin \left( \frac{\pi}{3} \right) $$

Since $ \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} $ and $ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $, the coordinates are:

$$ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $$

Find the angle θ in radians where cos(θ) = -1/2 and 0 ≤ θ < 2π

To find the angle $ \theta $ in radians where $ \cos(\theta) = -\frac{1}{2} $ and $ 0 \leq \theta < 2\pi $, we look for the points on the unit circle where the x-coordinate is -1/2.

These points correspond to angles in the second and third quadrants.

In the second quadrant, the angle is:

$$ \theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $$

In the third quadrant, the angle is:

$$ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $$

Therefore, the angles are:

$$ \theta = \frac{2\pi}{3} \text{ and } \frac{4\pi}{3} $$

Find the general solution of the equation tan(x) = cot(2x) on the unit circle

To solve the equation $ \tan(x) = \cot(2x) $ on the unit circle, we start by expressing $ \cot(2x) $ in terms of $ \tan(2x) $:

$$ \cot(2x) = \frac{1}{\tan(2x)} $$

The equation becomes:

$$ \tan(x) = \frac{1}{\tan(2x)} $$

Using the double-angle identity for tangent:

$$ \tan(2x) = \frac{2 \tan(x)}{1 – \tan^2(x)} $$

Substitute this back into the equation:

$$ \tan(x) = \frac{1}{\frac{2 \tan(x)}{1 – \tan^2(x)}} $$

Simplify the equation:

$$ \tan(x) = \frac{1 – \tan^2(x)}{2 \tan(x)} $$

Rearrange the equation:

$$ 2 \tan^2(x) = 1 – \tan^2(x) $$

Combine like terms:

$$ 3 \tan^2(x) = 1 $$

Solve for $ \tan(x) $:

$$ \tan(x) = \pm \frac{1}{\sqrt{3}} $$

Therefore, the general solution on the unit circle is:

$$ x = n\pi + (-1)^n \frac{\pi}{6} $$ where $ n $ is an integer.

Find the coordinates of the point on the unit circle corresponding to an angle of \( \frac{\pi}{4} \)

To find the coordinates of the point on the unit circle corresponding to an angle of $ \frac{\pi}{4} $, we use the unit circle definition:

The unit circle equation is: $$ x^2 + y^2 = 1 $$

At an angle of $ \frac{\pi}{4} $, both cosine and sine values are equal. Hence:

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

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

So, the coordinates are: $$ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$

Determine the coordinates of a point on the unit circle at an angle of 11π/6

To determine the coordinates of a point on the unit circle at an angle of $$ \frac{11\pi}{6} $$, we use the unit circle properties.

The coordinates can be found using the cosine and sine of the angle:

$$ x = \cos \left( \frac{11\pi}{6} \right) $$

$$ y = \sin \left( \frac{11\pi}{6} \right) $$

Since $$ \frac{11\pi}{6} $$ is in the fourth quadrant, we know:

$$ \cos \left( \frac{11\pi}{6} \right) = \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} $$

$$ \sin \left( \frac{11\pi}{6} \right) = -\sin \left( \frac{\pi}{6} \right) = -\frac{1}{2} $$

So the coordinates are:

$$ \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $$

Solve for the exact values of all angles θ in the interval [0, 2π) that satisfy cos(θ) = -1/2

To find the exact values of all angles $ \theta $ in the interval $ [0, 2\pi) $ that satisfy $ \cos(\theta) = -\frac{1}{2} $, we use the unit circle. The cosine value of $ -\frac{1}{2} $ corresponds to angles in the second and third quadrants. The reference angle is $ \frac{\pi}{3} $.

The angles are:

• In the second quadrant: $ \pi – \frac{\pi}{3} = \frac{2\pi}{3} $
• In the third quadrant: $ \pi + \frac{\pi}{3} = \frac{4\pi}{3} $

Thus, the solutions are:

$$ \theta = \frac{2\pi}{3}, \frac{4\pi}{3} $$

Determine the coordinates of the point where the terminal side of an angle of 5π/3 radians intersects the unit circle, and identify its quadrant

The angle $ \frac{5\pi}{3} $ radians is equivalent to 300 degrees (since $ \frac{5\pi}{3} \times \frac{180}{\pi} = 300 $ degrees).

This angle places the terminal side in the fourth quadrant.

In the fourth quadrant, the coordinates on the unit circle corresponding to an angle of 300 degrees are:

$$ ( \cos(300\degree), \sin(300\degree) ) $$

Since $ \cos(300\degree) = \cos(-60\degree) = \frac{1}{2} $ and $ \sin(300\degree) = \sin(-60\degree) = -\frac{\sqrt{3}}{2} $, the coordinates are:

$$ \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) $$

Thus, the terminal side intersects the unit circle at $ \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) $ in the fourth quadrant.

Find the values of tan(x) at different positions on the unit circle

To find the values of $ \tan(x) $ at different positions on the unit circle, we use the definition:

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

First, let

Calculate the integral of 1/(x + sqrt(x^2 – 1)) over the interval [-1, 1]

To calculate the integral:

$$ \int_{-1}^{1} \frac{1}{x + \sqrt{x^2 – 1}} dx $$

First, consider the substitution $ x = \cosh(t) $, which implies $ dx = \sinh(t) dt $.

When $ x = -1 $, $ t = i \pi $ and when $ x = 1 $, $ t = 0 $:

$$ \int_{i \pi}^{0} \frac{1}{\cosh(t) + \sinh(t)} \sinh(t) dt $$

Knowing that $ \cosh(t) + \sinh(t) = e^t $, the integral becomes:

$$ \int_{i \pi}^{0} \frac{\sinh(t)}{e^t} dt = \int_{i \pi}^{0} e^{-t} dt $$

Evaluating this gives:

$$ [ -e^{-t} ]_{i \pi}^{0} = -e^{0} + e^{-i \pi} = -1 + (-1) = -2 $$