QUESTION 2If Cos ⁡ Β = Ρ 5 \cos \beta = \frac{\rho}{\sqrt{5}} Cos Β = 5 ​ Ρ ​ Where Ρ \textgreater 0 \rho \ \textgreater \ 0 Ρ \textgreater 0 And $\beta \in [180^{\circ}, 360^{\circ}], Determine The Following Using A Diagram And Expressions In Terms Of Ρ \rho Ρ :1. $\tan

Introduction

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. In this article, we will focus on solving trigonometric equations, specifically the equation $\cos \beta = \frac{\rho}{\sqrt{5}}$, where $\rho > 0$ and $\beta \in [180^{\circ}, 360^{\circ}]$. We will use a diagram and expressions in terms of $\rho$ to determine the value of $\tan \beta$.

Understanding the Problem

The given equation is $\cos \beta = \frac{\rho}{\sqrt{5}}$. We are asked to find the value of $\tan \beta$ using a diagram and expressions in terms of $\rho$. To solve this problem, we need to understand the properties of the cosine function and how it relates to the tangent function.

Recall of Trigonometric Identities

Before we proceed, let's recall some important trigonometric identities:

• $\cos^2 \beta + \sin^2 \beta = 1$
• $\tan \beta = \frac{\sin \beta}{\cos \beta}$

These identities will be useful in solving the given equation.

Solving the Equation

To solve the equation $\cos \beta = \frac{\rho}{\sqrt{5}}$, we can use the following steps:

1. Square both sides: Square both sides of the equation to get rid of the square root.
2. Use the Pythagorean identity: Use the Pythagorean identity $\cos^2 \beta + \sin^2 \beta = 1$ to express $\sin^2 \beta$ in terms of $\cos^2 \beta$.
3. Solve for $\sin \beta$: Solve for $\sin \beta$ using the expression obtained in step 2.
4. Find $\tan \beta$: Find $\tan \beta$ using the expression for $\sin \beta$ and the identity $\tan \beta = \frac{\sin \beta}{\cos \beta}$.

Let's apply these steps to the given equation.

Step 1: Square both sides

$\cos^2 \beta = \left(\frac{\rho}{\sqrt{5}}\right)^2$

Step 2: Use the Pythagorean identity

$\cos^2 \beta + \sin^2 \beta = 1$

Substitute the expression for $\cos^2 \beta$ from step 1:

$\left(\frac{\rho}{\sqrt{5}}\right)^2 + \sin^2 \beta = 1$

Step 3: Solve for $\sin \beta$

$\sin^2 \beta = 1 - \left(\frac{\rho}{\sqrt{5}}\right)^2$

$\sin^2 \beta = 1 - \frac{\rho^2}{5}$

$\sin^2 \beta = \frac{5 - \rho^2}{5}$

Take the square root of both sides:

$\sin \beta = \pm \sqrt{\frac{5 - \rho^2}{5}}$

Step 4: Find $\tan \beta$

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

Substitute the expressions for $\sin \beta$ and $\cos \beta$:

$\tan \beta = \frac{\pm \sqrt{\frac{5 - \rho^2}{5}}}{\frac{\rho}{\sqrt{5}}}$

Simplify the expression:

$\tan \beta = \pm \sqrt{\frac{5 - \rho^2}{5}} \cdot \frac{\sqrt{5}}{\rho}$

$\tan \beta = \pm \frac{\sqrt{5 - \rho^2}}{\rho}$

Conclusion

In this article, we solved the trigonometric equation $\cos \beta = \frac{\rho}{\sqrt{5}}$ using a diagram and expressions in terms of $\rho$. We used the Pythagorean identity and the definition of the tangent function to find the value of $\tan \beta$. The final expression for $\tan \beta$ is $\pm \frac{\sqrt{5 - \rho^2}}{\rho}$.

Diagram

Here is a diagram that illustrates the solution:

+---------------+
|              |
|  cos β = ρ/√5  |
|              |
+---------------+
|  sin β = ±√(5-ρ^2)/5  |
|              |
+---------------+
|  tan β = ±√(5-ρ^2)/ρ  |
|              |
+---------------+

$$$$

Q&A: Frequently Asked Questions

Q: What is the main goal of solving trigonometric equations? A: The main goal of solving trigonometric equations is to find the value of an unknown angle or expression in terms of known values.

Q: What are some common trigonometric identities used to solve equations? A: Some common trigonometric identities used to solve equations include the Pythagorean identity ($\cos^2 \beta + \sin^2 \beta = 1$), the definition of the tangent function ($\tan \beta = \frac{\sin \beta}{\cos \beta}$), and the definition of the sine and cosine functions.

Q: How do you solve a trigonometric equation with a square root? A: To solve a trigonometric equation with a square root, you can use the following steps:

1. Square both sides of the equation to get rid of the square root.
2. Use the Pythagorean identity to express $\sin^2 \beta$ in terms of $\cos^2 \beta$.
3. Solve for $\sin \beta$ using the expression obtained in step 2.
4. Find $\tan \beta$ using the expression for $\sin \beta$ and the definition of the tangent function.

Q: What is the difference between the sine and cosine functions? A: The sine and cosine functions are both trigonometric functions that describe the relationships between the sides and angles of a right triangle. The sine function is defined as the ratio of the length of the opposite side to the length of the hypotenuse, while the cosine function is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.

Q: How do you find the value of $\tan \beta$ in terms of $\rho$? A: To find the value of $\tan \beta$ in terms of $\rho$, you can use the following steps:

1. Square both sides of the equation $\cos \beta = \frac{\rho}{\sqrt{5}}$ to get rid of the square root.
2. Use the Pythagorean identity to express $\sin^2 \beta$ in terms of $\cos^2 \beta$.
3. Solve for $\sin \beta$ using the expression obtained in step 2.
4. Find $\tan \beta$ using the expression for $\sin \beta$ and the definition of the tangent function.

Q: What is the final expression for $\tan \beta$ in terms of $\rho$? A: The final expression for $\tan \beta$ in terms of $\rho$ is $\pm \frac{\sqrt{5 - \rho^2}}{\rho}$.

Q: Can you provide a diagram that illustrates the solution? A: Yes, here is a diagram that illustrates the solution:

+---------------+
|              |
|  cos β = ρ/√5  |
|              |
+---------------+
|  sin β = ±√(5-ρ^2)/5  |
|              |
+---------------+
|  tan β = ±√(5-ρ^2)/ρ  |
|              |
+---------------+

This diagram shows the relationships between the cosine, sine, and tangent functions, as well as the expression for $\tan \beta$ in terms of $\rho$.

Conclusion

In this article, we provided a step-by-step guide to solving trigonometric equations, including the equation $\cos \beta = \frac{\rho}{\sqrt{5}}$. We also answered some frequently asked questions about trigonometric equations and provided a diagram that illustrates the solution. We hope this article has been helpful in understanding the concepts of trigonometric equations and how to solve them.