Let \[$\sin A = -\frac{3}{5}\$\] With \[$A\$\] In QIV, And \[$\cos B = -\frac{12}{13}\$\] With \[$B\$\] In QII. Answer Exactly.a. Find \[$\sin(A-B)\$\].\[$\square\$\]b. Find

by ADMIN 174 views

Introduction

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is a fundamental subject that has numerous applications in various fields, including physics, engineering, and navigation. In this article, we will explore some of the key concepts in trigonometry, including the sine and cosine functions, and how to use them to solve problems involving angles and triangles.

The Sine and Cosine Functions

The sine and cosine functions are two of the most important trigonometric functions. The sine function, denoted by sin(x), is defined as the ratio of the length of the side opposite the angle x to the length of the hypotenuse (the side opposite the right angle). The cosine function, denoted by cos(x), is defined as the ratio of the length of the side adjacent to the angle x to the length of the hypotenuse.

Given Information

We are given two pieces of information:

  • sin⁑A=βˆ’35\sin A = -\frac{3}{5}, where A is an angle in the fourth quadrant (QIV)
  • cos⁑B=βˆ’1213\cos B = -\frac{12}{13}, where B is an angle in the second quadrant (QII)

Finding Sin(A-B)

To find sin⁑(Aβˆ’B)\sin(A-B), we can use the following formula:

sin⁑(Aβˆ’B)=sin⁑Acos⁑Bβˆ’cos⁑Asin⁑B\sin(A-B) = \sin A \cos B - \cos A \sin B

However, we are not given the values of cos⁑A\cos A and sin⁑B\sin B. We can use the Pythagorean identity to find these values.

The Pythagorean Identity

The Pythagorean identity states that:

sin⁑2x+cos⁑2x=1\sin^2 x + \cos^2 x = 1

We can use this identity to find the values of cos⁑A\cos A and sin⁑B\sin B.

Finding Cos A

We are given that sin⁑A=βˆ’35\sin A = -\frac{3}{5}. We can use the Pythagorean identity to find the value of cos⁑A\cos A.

cos⁑2A=1βˆ’sin⁑2A\cos^2 A = 1 - \sin^2 A

cos⁑2A=1βˆ’(βˆ’35)2\cos^2 A = 1 - \left(-\frac{3}{5}\right)^2

cos⁑2A=1βˆ’925\cos^2 A = 1 - \frac{9}{25}

cos⁑2A=1625\cos^2 A = \frac{16}{25}

cos⁑A=±1625\cos A = \pm \sqrt{\frac{16}{25}}

Since A is an angle in the fourth quadrant, cos⁑A\cos A is negative.

cos⁑A=βˆ’1625\cos A = -\sqrt{\frac{16}{25}}

cos⁑A=βˆ’45\cos A = -\frac{4}{5}

Finding Sin B

We are given that cos⁑B=βˆ’1213\cos B = -\frac{12}{13}. We can use the Pythagorean identity to find the value of sin⁑B\sin B.

sin⁑2B=1βˆ’cos⁑2B\sin^2 B = 1 - \cos^2 B

sin⁑2B=1βˆ’(βˆ’1213)2\sin^2 B = 1 - \left(-\frac{12}{13}\right)^2

sin⁑2B=1βˆ’144169\sin^2 B = 1 - \frac{144}{169}

sin⁑2B=25169\sin^2 B = \frac{25}{169}

sin⁑B=±25169\sin B = \pm \sqrt{\frac{25}{169}}

Since B is an angle in the second quadrant, sin⁑B\sin B is positive.

sin⁑B=25169\sin B = \sqrt{\frac{25}{169}}

sin⁑B=513\sin B = \frac{5}{13}

Finding Sin(A-B)

Now that we have the values of cos⁑A\cos A, sin⁑A\sin A, cos⁑B\cos B, and sin⁑B\sin B, we can find the value of sin⁑(Aβˆ’B)\sin(A-B).

sin⁑(Aβˆ’B)=sin⁑Acos⁑Bβˆ’cos⁑Asin⁑B\sin(A-B) = \sin A \cos B - \cos A \sin B

sin⁑(Aβˆ’B)=(βˆ’35)(βˆ’1213)βˆ’(βˆ’45)(513)\sin(A-B) = \left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right) - \left(-\frac{4}{5}\right)\left(\frac{5}{13}\right)

sin⁑(Aβˆ’B)=3665+2065\sin(A-B) = \frac{36}{65} + \frac{20}{65}

sin⁑(Aβˆ’B)=5665\sin(A-B) = \frac{56}{65}

Conclusion

In this article, we used the sine and cosine functions to solve a problem involving angles and triangles. We found the values of cos⁑A\cos A and sin⁑B\sin B using the Pythagorean identity, and then used these values to find the value of sin⁑(Aβˆ’B)\sin(A-B). This problem demonstrates the importance of trigonometry in solving real-world problems.

Final Answer

The final answer is 5665\boxed{\frac{56}{65}}.

Additional Problems

Problem 1

Find cos⁑(Aβˆ’B)\cos(A-B).

Solution

We can use the following formula:

cos⁑(Aβˆ’B)=cos⁑Acos⁑B+sin⁑Asin⁑B\cos(A-B) = \cos A \cos B + \sin A \sin B

We are given that cos⁑A=βˆ’45\cos A = -\frac{4}{5}, sin⁑A=βˆ’35\sin A = -\frac{3}{5}, cos⁑B=βˆ’1213\cos B = -\frac{12}{13}, and sin⁑B=513\sin B = \frac{5}{13}.

cos⁑(Aβˆ’B)=(βˆ’45)(βˆ’1213)+(βˆ’35)(513)\cos(A-B) = \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) + \left(-\frac{3}{5}\right)\left(\frac{5}{13}\right)

cos⁑(Aβˆ’B)=4865βˆ’1565\cos(A-B) = \frac{48}{65} - \frac{15}{65}

cos⁑(Aβˆ’B)=3365\cos(A-B) = \frac{33}{65}

Problem 2

Find tan⁑(Aβˆ’B)\tan(A-B).

Solution

We can use the following formula:

tan⁑(Aβˆ’B)=tan⁑Aβˆ’tan⁑B1+tan⁑Atan⁑B\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

We are given that tan⁑A=sin⁑Acos⁑A=βˆ’35βˆ’45=34\tan A = \frac{\sin A}{\cos A} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4} and tan⁑B=sin⁑Bcos⁑B=513βˆ’1213=βˆ’512\tan B = \frac{\sin B}{\cos B} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12}.

tan⁑(Aβˆ’B)=34βˆ’(βˆ’512)1+(34)(βˆ’512)\tan(A-B) = \frac{\frac{3}{4} - \left(-\frac{5}{12}\right)}{1 + \left(\frac{3}{4}\right)\left(-\frac{5}{12}\right)}

tan⁑(Aβˆ’B)=912+5121βˆ’1548\tan(A-B) = \frac{\frac{9}{12} + \frac{5}{12}}{1 - \frac{15}{48}}

tan⁑(Aβˆ’B)=14123348\tan(A-B) = \frac{\frac{14}{12}}{\frac{33}{48}}

tan⁑(Aβˆ’B)=1412Γ—4833\tan(A-B) = \frac{14}{12} \times \frac{48}{33}

tan⁑(Aβˆ’B)=672396\tan(A-B) = \frac{672}{396}

\tan(A-B) = \frac{56}{33}$<br/> **Q&A: Trigonometry and Beyond** ============================= **Q: What is Trigonometry?** ------------------------- A: Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is a fundamental subject that has numerous applications in various fields, including physics, engineering, and navigation. **Q: What are the Sine and Cosine Functions?** ----------------------------------------- A: The sine and cosine functions are two of the most important trigonometric functions. The sine function, denoted by sin(x), is defined as the ratio of the length of the side opposite the angle x to the length of the hypotenuse (the side opposite the right angle). The cosine function, denoted by cos(x), is defined as the ratio of the length of the side adjacent to the angle x to the length of the hypotenuse. **Q: How do I Use the Pythagorean Identity?** ----------------------------------------- A: The Pythagorean identity states that: $\sin^2 x + \cos^2 x = 1

You can use this identity to find the values of cos⁑A\cos A and sin⁑B\sin B when given the values of sin⁑A\sin A and cos⁑B\cos B.

Q: How do I Find Sin(A-B)?

A: To find sin⁑(Aβˆ’B)\sin(A-B), you can use the following formula:

sin⁑(Aβˆ’B)=sin⁑Acos⁑Bβˆ’cos⁑Asin⁑B\sin(A-B) = \sin A \cos B - \cos A \sin B

You will need to find the values of cos⁑A\cos A and sin⁑B\sin B using the Pythagorean identity.

Q: How do I Find Cos(A-B)?

A: To find cos⁑(Aβˆ’B)\cos(A-B), you can use the following formula:

cos⁑(Aβˆ’B)=cos⁑Acos⁑B+sin⁑Asin⁑B\cos(A-B) = \cos A \cos B + \sin A \sin B

You will need to find the values of cos⁑A\cos A and sin⁑B\sin B using the Pythagorean identity.

Q: How do I Find Tan(A-B)?

A: To find tan⁑(Aβˆ’B)\tan(A-B), you can use the following formula:

tan⁑(Aβˆ’B)=tan⁑Aβˆ’tan⁑B1+tan⁑Atan⁑B\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

You will need to find the values of tan⁑A\tan A and tan⁑B\tan B using the given values of sin⁑A\sin A, cos⁑A\cos A, sin⁑B\sin B, and cos⁑B\cos B.

Q: What are Some Common Trigonometric Identities?

A: Some common trigonometric identities include:

  • sin⁑2x+cos⁑2x=1\sin^2 x + \cos^2 x = 1
  • tan⁑x=sin⁑xcos⁑x\tan x = \frac{\sin x}{\cos x}
  • cot⁑x=cos⁑xsin⁑x\cot x = \frac{\cos x}{\sin x}
  • sec⁑x=1cos⁑x\sec x = \frac{1}{\cos x}
  • csc⁑x=1sin⁑x\csc x = \frac{1}{\sin x}

Q: How do I Use Trigonometry in Real-World Applications?

A: Trigonometry has numerous applications in various fields, including:

  • Physics: Trigonometry is used to describe the motion of objects in terms of position, velocity, and acceleration.
  • Engineering: Trigonometry is used to design and build structures, such as bridges and buildings.
  • Navigation: Trigonometry is used to determine the position and direction of objects, such as ships and planes.
  • Computer Science: Trigonometry is used in computer graphics and game development to create 3D models and animations.

Q: What are Some Common Mistakes to Avoid in Trigonometry?

A: Some common mistakes to avoid in trigonometry include:

  • Confusing the sine and cosine functions.
  • Failing to use the Pythagorean identity to find the values of cos⁑A\cos A and sin⁑B\sin B.
  • Making errors when using the formulas for sin⁑(Aβˆ’B)\sin(A-B), cos⁑(Aβˆ’B)\cos(A-B), and tan⁑(Aβˆ’B)\tan(A-B).
  • Failing to check the units and dimensions of the given values.

Conclusion

Trigonometry is a fundamental subject that has numerous applications in various fields. By understanding the sine and cosine functions, the Pythagorean identity, and the formulas for sin⁑(Aβˆ’B)\sin(A-B), cos⁑(Aβˆ’B)\cos(A-B), and tan⁑(Aβˆ’B)\tan(A-B), you can solve a wide range of problems involving angles and triangles. Remember to avoid common mistakes and to use trigonometry in real-world applications.

Final Answer

The final answer is 5665\boxed{\frac{56}{65}}.