Math

Problem: Calculate the Sine, Cosine, and Tangent Values of Specific Angles on the Unit Circle

Let’s determine the sine, cosine, and tangent values for the angle θ = 225° on the unit circle.

First, convert the angle to radians:

$$ θ = 225° = \frac{225π}{180} = \frac{5π}{4} radians $$

Using the properties of the unit circle, we know:

$$ \cos(\frac{5π}{4}) = -\frac{\sqrt{2}}{2} $$

$$ \sin(\frac{5π}{4}) = -\frac{\sqrt{2}}{2} $$

$$ \tan(\frac{5π}{4}) = \frac{\sin(\frac{5π}{4})}{\cos(\frac{5π}{4})} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1 $$

Thus, the sine, cosine, and tangent values for θ = 225° are:

$$ \sin(225°) = -\frac{\sqrt{2}}{2} $$

$$ \cos(225°) = -\frac{\sqrt{2}}{2} $$

$$ \tan(225°) = 1 $$

Calculating the Tangent Value of an Angle in the Unit Circle

Let’s consider an angle $ \theta $ in the unit circle. The coordinates of a point on the unit circle are given by $(\cos \theta, \sin \theta)$. The tangent of the angle $ \theta $ is defined as:

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

Suppose $\theta = \frac{5\pi}{4}$, we need to find the value of $\tan \theta$. From the unit circle, we have:

$\sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$

$\cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$

Thus,

$$\tan \frac{5\pi}{4} = \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1$$

Determine the tangent values for specific angles on the unit circle

To determine the tangent values for angles $\frac{\pi}{4}$, $\frac{2\pi}{3}$, and $\frac{5\pi}{6}$ on the unit circle, follow these steps:

1. For the angle $\frac{\pi}{4}$: $$\tan \left( \frac{\pi}{4} \right) = 1$$

2. For the angle $\frac{2\pi}{3}$: $$\tan \left( \frac{2\pi}{3} \right) = -\sqrt{3}$$

3. For the angle $\frac{5\pi}{6}$: $$\tan \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{3}$$

Thus, the tangent values are $1$, $-\sqrt{3}$, and $-\frac{\sqrt{3}}{3}$ respectively.

Given that the angle θ in standard position intersects the unit circle at the point (x, y) in the first quadrant where x = 3/5, find the y-coordinate of the point Use the Pythagorean identity for the unit circle to show your work

Given the Pythagorean identity for the unit circle:

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

where $$ x = \frac{3}{5}$$, substitute this value into the identity:

$$ \left( \frac{3}{5} \right)^2 + y^2 = 1 $$

$$ \frac{9}{25} + y^2 = 1 $$

Subtract $$ \frac{9}{25}$$ from both sides:

$$ y^2 = 1 – \frac{9}{25} $$

$$ y^2 = \frac{25}{25} – \frac{9}{25} $$

$$ y^2 = \frac{16}{25} $$

Taking the square root of both sides:

$$ y = \pm \sqrt{\frac{16}{25}} $$

$$ y = \pm \frac{4}{5} $$

Since (x, y) is in the first quadrant:

$$ y = \frac{4}{5} $$

Find the angle whose cosine is -2/3 using the unit circle

To find the angle whose cosine is $-\frac{2}{3}$, we need to look at the unit circle and identify the angles where the x-coordinate (cosine value) is $-\frac{2}{3}$. Since cosine is negative in the second and third quadrants, we look in those regions.

Thus, we have:

$$\theta = \cos^{-1}(-\frac{2}{3})$$

and

$$\theta = 2\pi – \cos^{-1}(-\frac{2}{3})$$

These angles in degrees are approximately:

$$\theta \approx 131.81^\circ$$

and

$$\theta \approx 228.19^\circ$$

Find the Cotangent of an Angle on the Unit Circle

To find the cotangent of an angle $\theta$ on the unit circle, we use the identity:

$$ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} $$

Given $\theta = \frac{3\pi}{4}$, we know from the unit circle that:

$$ \cos \left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2} $$

and

$$ \sin \left( \frac{3\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

Therefore,

$$ \cot \left( \frac{3\pi}{4} \right) = \frac{\cos \left( \frac{3\pi}{4} \right)}{\sin \left( \frac{3\pi}{4} \right)} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 $$

So, the cotangent of $\frac{3\pi}{4}$ is $-1$.

Find the angle where tan(θ) = -1 in the unit circle

To find the angle where $\tan(\theta) = -1$ in the unit circle, we need to look for the values of $\theta$ where the tangent function is negative and equals -1.

We know that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. For $\tan(\theta) = -1$, this means $\sin(\theta) = -\cos(\theta)$.

This occurs in the second and fourth quadrants.

In the second quadrant: $\theta = \pi – \frac{\pi}{4} = \frac{3\pi}{4}$

In the fourth quadrant: $\theta = 2\pi – \frac{\pi}{4} = \frac{7\pi}{4}$

Hence, the angles are $\theta = \frac{3\pi}{4}$ and $\theta = \frac{7\pi}{4}$.

Determine the exact values of trigonometric functions at specific angles using the unit circle

To find the exact values of trigonometric functions for $ \theta = \frac{5\pi}{6} $, we first recognize that this angle corresponds to a reference angle of $ \frac{\pi}{6} $ in the second quadrant.

The coordinates of the point on the unit circle at $ \frac{\pi}{6} $ are $ (\frac{\sqrt{3}}{2}, \frac{1}{2}) $. Since $ \frac{5\pi}{6} $ lies in the second quadrant, the x-coordinate becomes negative:

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

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

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