Unit Circle

Find the sine and cosine of the angle θ = 45° using the unit circle

To find the sine and cosine of the angle $\theta = 45°$, we use the unit circle where the radius is 1.

The coordinates of the point where the terminal side of a 45° angle intersects the unit circle are $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

Therefore, $$\cos(45°) = \frac{\sqrt{2}}{2}$$ and $$\sin(45°) = \frac{\sqrt{2}}{2}$$.

Determine the Values of Trigonometric Functions at π/3

Consider the angle $\frac{\pi}{3}$ on the unit circle. To find the values of sin, cos, and tan at this angle, we use the known values:

The sine of $\frac{\pi}{3}$ is:

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

The cosine of $\frac{\pi}{3}$ is:

$$\cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$$

Using the quotient identity for tangent:

$$\tan \left( \frac{\pi}{3} \right) = \frac{\sin \left( \frac{\pi}{3} \right)}{\cos \left( \frac{\pi}{3} \right)} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$$

Determine the cosine value of -π/3 using the unit circle

First, recall that the unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. In the unit circle, the angle $\theta = -\pi/3$ is measured in the clockwise direction.

To find the cosine of $-\pi/3$, we can use the symmetry of the unit circle. The angle $-\pi/3$ is the same as $5\pi/3$ in the standard position (i.e., measured counterclockwise from the positive x-axis).

Cosine corresponds to the x-coordinate of the point on the unit circle. Thus, we need to find the x-coordinate of the point corresponding to $5\pi/3$.

At $5\pi/3$, the point on the unit circle is $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$. Therefore, the cosine of $-\pi/3$ is:

$$\cos(-\pi/3) = \frac{1}{2}$$

Find the value of tan(-π/6) using the unit circle

We start by recognizing that the angle $-\frac{\pi}{6}$ is equivalent to rotating $\frac{\pi}{6}$ radians in the clockwise direction.

On the unit circle, the point corresponding to $\frac{\pi}{6}$ radians is $(\frac{\sqrt{3}}{2}, \frac{1}{2})$. When we rotate in the clockwise direction to $-\frac{\pi}{6}$, the coordinates of the point become $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$.

The tangent of an angle is given by the ratio of the y-coordinate to the x-coordinate:

$$\tan(-\frac{\pi}{6}) = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$$

So, the value of $\tan(-\frac{\pi}{6})$ is $-\frac{\sqrt{3}}{3}$.

Understanding the representation of sine on the unit circle

To understand what sine represents on the unit circle, let’s begin with the definition of the unit circle. The unit circle is a circle with a radius of 1 centered at the origin of the coordinate system.

Consider a point $P(x, y)$ on the unit circle that forms an angle $\theta$ with the positive x-axis. The coordinates of point $P$ can be expressed in terms of trigonometric functions as:

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

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

Therefore, the sine of the angle $\theta$ is the y-coordinate of the corresponding point on the unit circle.

To elaborate with a specific angle, let’s consider $\theta = \frac{\pi}{4}$. The coordinates of the point on the unit circle at this angle are:

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

Thus, for $\theta = \frac{\pi}{4}$, the sine value is:

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

This demonstrates that sine represents the y-coordinate of a point on the unit circle corresponding to a given angle.

Given a point on the unit circle at an angle theta in radians, determine the exact coordinates and verify their correctness for theta = 7π/6

We start with the unit circle formula:

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

Given $\theta = \frac{7\pi}{6}$, we need to find the cosine and sine of this angle:

$$\cos\left( \frac{7\pi}{6} \right) = \cos\left( \pi + \frac{\pi}{6} \right) = -\cos\left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2}$$

$$\sin\left( \frac{7\pi}{6} \right) = \sin\left( \pi + \frac{\pi}{6} \right) = -\sin\left( \frac{\pi}{6} \right) = -\frac{1}{2}$$

Thus, the coordinates are:

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

Verification:

$$\left( -\frac{\sqrt{3}}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 = \frac{3}{4} + \frac{1}{4} = 1$$

The coordinates are correct.

Given a point P on the unit circle with coordinates (x, y), find the value of cotangent at angle θ where θ is the angle formed by the positive x-axis and the line segment OP

Given a point $P(x, y)$ on the unit circle:

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

The cotangent of angle $\theta$ is:

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

Since $x = \cos(\theta)$ and $y = \sin(\theta)$, we can write:

$$\cot(\theta) = \frac{x}{y}$$

Therefore, the value of $\cot(\theta)$ is:

$$\cot(\theta) = \frac{x}{y}$$

Finding the Coordinates of a Point on the Unit Circle Given an Angle

Given an angle \( \theta \) in radians, find the coordinates of the corresponding point on the unit circle.

Step 1: Recall the unit circle definition. The unit circle is a circle with a radius of 1 centered at the origin \((0, 0)\) in the coordinate plane.

Step 2: Use the cosine and sine functions to determine the coordinates: \(x = \cos(\theta)\), \(y = \sin(\theta)\).

Step 3: Compute the coordinates for a given angle \( \theta = \frac{5\pi}{6} \).

$$x = \cos \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{2}$$

$$y = \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2}$$

Therefore, the coordinates are \( \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).

Find the Intersection Points on the Unit Circle

Consider the unit circle defined by the equation $x^2 + y^2 = 1$ and the line with the equation $y = mx + c$.

To find the intersection points, substitute $y = mx + c$ into the unit circle’s equation:

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

Expand and simplify:

$$x^2 + m^2x^2 + 2mxc + c^2 = 1$$

Combine like terms:

$$ (1 + m^2)x^2 + 2mxc + c^2 – 1 = 0 $$

This is a quadratic equation in $x$:

$$ Ax^2 + Bx + C = 0 $$

where $A = 1 + m^2$, $B = 2mc$, and $C = c^2 – 1$.

Solve for $x$ using the quadratic formula:

$$ x = \frac{-B \pm \sqrt{B^2 – 4AC}}{2A} $$

Substitute the values of $A$, $B$, and $C$:

$$ x = \frac{-2mc \pm \sqrt{(2mc)^2 – 4(1 + m^2)(c^2 – 1)}}{2(1 + m^2)} $$

Simplify further to find $x$, then use $y = mx + c$ to find the $y$ values of the intersection points.

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

For an angle \(\theta = \frac{\pi}{4}\) on the unit circle, we use the trigonometric functions sine and cosine to find the coordinates. The coordinates are given by \((\cos \theta, \sin \theta)\).

$$ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

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

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