Math

Find the value of sin(θ) for θ = 7π/6 using the unit circle

To find $\sin(\theta)$ for $\theta = \frac{7\pi}{6}$, we need to locate the angle on the unit circle.

First, note that $\frac{7\pi}{6}$ is in the third quadrant where sine is negative.

$\frac{7\pi}{6}$ is $30^\circ$ past $\pi$ (180 degrees).

The reference angle is $30^\circ$ or $\frac{\pi}{6}$.

In the third quadrant, the sine of $\frac{\pi}{6}$ is $-\frac{1}{2}$.

Thus, $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$.

Determine Specific Coordinates on the Unit Circle

Given a point on the unit circle defined by the angle $\theta$ in radians, find the coordinates of the point where $\theta = \frac{5\pi}{6}$.

First, recall that any point on the unit circle is defined by $(\cos \theta, \sin \theta)$. Here $\theta = \frac{5\pi}{6}$.

So, we need to find $\cos \left(\frac{5\pi}{6}\right)$ and $\sin \left(\frac{5\pi}{6}\right)$:

• For $\cos \left(\frac{5\pi}{6}\right)$: Since $\frac{5\pi}{6}$ is in the second quadrant, $\cos \left(\frac{5\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
• For $\sin \left(\frac{5\pi}{6}\right)$: $\sin$ is positive in the second quadrant, so $\sin \left(\frac{5\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Thus, the coordinates for $\theta = \frac{5\pi}{6}$ are $$\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).$$

Find the sine, cosine, and tangent values for the angle $\theta = \frac{\pi}{4}$ on the unit circle

To find the sine, cosine, and tangent values for $\theta = \frac{\pi}{4}$, we use the unit circle properties.

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

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

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

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

Find the tangent line to the unit circle at the point (1, 0)

The equation of the unit circle is $x^2 + y^2 = 1$.

The point (1,0) is on the unit circle.

The slope of the tangent line at any point (a,b) on the circle can be found by implicit differentiation:

$$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = – \frac{x}{y}$$

At the point (1,0), the slope is undefined (vertical line).

Therefore, the equation of the tangent line is $$ x = 1 $$

Identify the quadrant of an angle

Given an angle of $135°$, determine the quadrant in which this angle lies on the unit circle.

The unit circle is divided into four quadrants:

Quadrant I: $0°$ to $90°$

Quadrant II: $90°$ to $180°$

Quadrant III: $180°$ to $270°$

Quadrant IV: $270°$ to $360°$

Since $135°$ is greater than $90°$ and less than $180°$,

$$135°$$

lies in Quadrant II.

Find the value of tan(θ) on the unit circle where θ is a special angle

To find the value of $\tan(\theta)$ on the unit circle, we need to use the relationship $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.

Let’s consider $\theta = \frac{\pi}{4}$. On the unit circle, $\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

Thus,

$$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1.$$

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

Find the coordinates on the Unit Circle

To determine the coordinates on the unit circle corresponding to an angle of $ \frac{5\pi}{6} $, we use the trigonometric functions sine and cosine.

The cosine of $ \frac{5\pi}{6} $ corresponds to the x-coordinate, and the sine of $ \frac{5\pi}{6} $ corresponds to the y-coordinate.

Calculating these values:

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

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

So, the coordinates are:

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

Find the exact value of the trigonometric functions for the angle θ = 5π/6 using the unit circle

First, locate the angle $\theta = \frac{5\pi}{6}$ on the unit circle. This angle is in the second quadrant.

In the second quadrant, sine is positive and cosine is negative.

The reference angle for $\theta = \frac{5\pi}{6}$ is $\frac{\pi}{6}$.

From the unit circle, $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.

Therefore, $\sin(\frac{5\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{5\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$.

Hence, the exact values are:

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

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

$$\tan(\frac{5\pi}{6}) = \frac{\sin(\frac{5\pi}{6})}{\cos(\frac{5\pi}{6})} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$

Find the values of tan(θ) for specific angles on the unit circle

$$For \ θ = \frac{3π}{4}, \ we \ know \ that \ tan(θ) = \frac{sin(θ)}{cos(θ)}$$

$$sin(θ) = sin(\frac{3π}{4}) = \frac{1}{\sqrt{2}}, \ cos(θ) = cos(\frac{3π}{4}) = -\frac{1}{\sqrt{2}}$$

$$Therefore, \ tan(θ) = \frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}} = -1$$

Find the value of tan for the angle 45 degrees on the unit circle

To find the value of $\tan 45^{\circ}$ on the unit circle, we use the definition of $\tan$:

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

For $\theta = 45^{\circ}$, we know that $\sin 45^{\circ} = \frac{\sqrt{2}}{2}$ and $\cos 45^{\circ} = \frac{\sqrt{2}}{2}$.

Substituting these values in, we get:

$$\tan 45^{\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$$

So, the value of $\tan 45^{\circ}$ is 1.