Math

Explain the concept of a unit circle, including its importance in trigonometry and how it relates to the coordinates of points on the circle

A unit circle is a circle with a radius of 1, centered at the origin of a coordinate system. The equation of the unit circle is given by:

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

The unit circle is fundamental in trigonometry as it defines the sine and cosine functions for all real numbers. For any angle $\theta$, the coordinates of the corresponding point on the unit circle are $(\cos(\theta), \sin(\theta))$. These coordinates are derived from the definitions:

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

$$ \sin(\theta) = \frac{y}{1} = y $$

Additionally, the unit circle helps in visualizing and understanding periodic properties of trigonometric functions and their symmetries.

Find the angle in degrees corresponding to 7π/6 radians on the unit circle

To convert $\frac{7\pi}{6}$ radians to degrees, we use the conversion factor:

$$ 180^{\circ} = \pi \text{ radians} $$

Thus,

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

The angle in degrees is:

$$ 210^{\circ} $$

Find the coordinates on the unit circle where the tangent of the angle is 1

To find the coordinates on the unit circle where $ \tan(\theta) = 1 $, we need to determine the angles $\theta $ for which this condition holds. We know that:

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

For the tangent to be 1, the sine and cosine must be equal. This occurs at angles:

$$ \theta = \frac {\pi}{4} \text{ and } \theta = \frac {5\pi}{4} $$

Now, we find the coordinates on the unit circle for these angles:

$$ \text{At } \theta = \frac {\pi}{4}, \text{ the coordinates are } \left( \frac {\sqrt {2}}{2}, \frac {\sqrt {2}}{2} \right) $$

$$ \text{At } \theta = \frac {5\pi}{4}, \text{ the coordinates are } \left( -\frac {\sqrt {2}}{2}, -\frac {\sqrt {2}}{2} \right) $$

Thus, the coordinates on the unit circle where $ \tan(\theta) = 1 $ are:

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

and

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

Find the sine value at π/3 on the unit circle

To find the sine value at $ \frac{\pi}{3} $ on the unit circle, we use the unit circle definition. The angle $ \frac{\pi}{3} $ is equivalent to 60 degrees.

On the unit circle, the sine value at 60 degrees is:

$$ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} $$

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:

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