Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To solve the equation [tex]\(-\frac{1}{\tan A} - \frac{1}{\tan 2A} = \frac{1}{\sin 2A}\)[/tex], we can proceed with the following steps:
1. Recall the trigonometric identities that we will need:
- [tex]\(\tan(2A) = \frac{2 \tan(A)}{1 - \tan^2(A)}\)[/tex]
- [tex]\(\sin(2A) = 2 \sin(A) \cos(A)\)[/tex]
2. Substitute these identities into the equation:
- Substitute [tex]\(\tan(2A)\)[/tex] and [tex]\(\sin(2A)\)[/tex] into the equation.
- First, we replace [tex]\(\tan(2A)\)[/tex]:
[tex]\[ \frac{1}{\tan(2A)} = \frac{1}{\frac{2 \tan(A)}{1 - \tan^2(A)}} = \frac{1 - \tan^2(A)}{2 \tan(A)} \][/tex]
- Substitute [tex]\(\sin(2A)\)[/tex]:
[tex]\[ \frac{1}{\sin(2A)} = \frac{1}{2 \sin(A) \cos(A)} \][/tex]
3. Rewrite the equation with the substitutions:
[tex]\[ -\frac{1}{\tan(A)} - \frac{1 - \tan^2(A)}{2 \tan(A)} = \frac{1}{2 \sin(A) \cos(A)} \][/tex]
4. Simplify the left-hand side:
- Combine the fractions on the left-hand side:
[tex]\[ -\frac{1}{\tan(A)} - \frac{1 - \tan^2(A)}{2 \tan(A)} = -\frac{2 - (1 - \tan^2(A))}{2 \tan(A)} = -\frac{1 + \tan^2(A)}{2 \tan(A)} \][/tex]
- Since [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex], [tex]\(\tan^2(A) = \frac{\sin^2(A)}{\cos^2(A)}\)[/tex], we use:
[tex]\[ -\frac{1 + \tan^2(A)}{2 \tan(A)} = -\frac{1 + \frac{\sin^2(A)}{\cos^2(A)}}{2 \cdot \frac{\sin(A)}{\cos(A)}} = -\frac{\cos^2(A) + \sin^2(A)}{2 \sin(A) \cos(A)} = -\frac{1}{2 \sin(A) \cos(A)} \][/tex]
5. Simplify the right-hand side:
[tex]\[ \frac{1}{2 \sin(A) \cos(A)} \][/tex]
6. Finally, compare the left-hand side and the right-hand side:
- Left-hand side:
[tex]\[ -\frac{1}{2 \sin(A) \cos(A)} \][/tex]
- Right-hand side:
[tex]\[ \frac{1}{2 \sin(A) \cos(A)} \][/tex]
We find that the simplified form of the left-hand side is:
[tex]\[ -\frac{1}{2 \sin(A) \cos(A)} - \frac{\sin(3A)}{\sin(A)} \cdot \frac{1}{2 \sin(A) \cos(A)} \][/tex]
Since these do not equate ([tex]\((1 + \sin(3A)/\sin(A))\)[/tex] is non-zero), we conclude that:
[tex]\[ -\frac{1}{\tan A} - \frac{1}{\tan 2A} \neq \frac{1}{\sin 2A} \][/tex]
So, the equation simplifies to:
[tex]\[ -(1 + \frac{\sin(3A)}{\sin(A)})/(2 \sin(A) \cos(A)) \neq 0 \][/tex]
This shows that the original equation does not simplify to equality for any general value of [tex]\(A\)[/tex].
1. Recall the trigonometric identities that we will need:
- [tex]\(\tan(2A) = \frac{2 \tan(A)}{1 - \tan^2(A)}\)[/tex]
- [tex]\(\sin(2A) = 2 \sin(A) \cos(A)\)[/tex]
2. Substitute these identities into the equation:
- Substitute [tex]\(\tan(2A)\)[/tex] and [tex]\(\sin(2A)\)[/tex] into the equation.
- First, we replace [tex]\(\tan(2A)\)[/tex]:
[tex]\[ \frac{1}{\tan(2A)} = \frac{1}{\frac{2 \tan(A)}{1 - \tan^2(A)}} = \frac{1 - \tan^2(A)}{2 \tan(A)} \][/tex]
- Substitute [tex]\(\sin(2A)\)[/tex]:
[tex]\[ \frac{1}{\sin(2A)} = \frac{1}{2 \sin(A) \cos(A)} \][/tex]
3. Rewrite the equation with the substitutions:
[tex]\[ -\frac{1}{\tan(A)} - \frac{1 - \tan^2(A)}{2 \tan(A)} = \frac{1}{2 \sin(A) \cos(A)} \][/tex]
4. Simplify the left-hand side:
- Combine the fractions on the left-hand side:
[tex]\[ -\frac{1}{\tan(A)} - \frac{1 - \tan^2(A)}{2 \tan(A)} = -\frac{2 - (1 - \tan^2(A))}{2 \tan(A)} = -\frac{1 + \tan^2(A)}{2 \tan(A)} \][/tex]
- Since [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex], [tex]\(\tan^2(A) = \frac{\sin^2(A)}{\cos^2(A)}\)[/tex], we use:
[tex]\[ -\frac{1 + \tan^2(A)}{2 \tan(A)} = -\frac{1 + \frac{\sin^2(A)}{\cos^2(A)}}{2 \cdot \frac{\sin(A)}{\cos(A)}} = -\frac{\cos^2(A) + \sin^2(A)}{2 \sin(A) \cos(A)} = -\frac{1}{2 \sin(A) \cos(A)} \][/tex]
5. Simplify the right-hand side:
[tex]\[ \frac{1}{2 \sin(A) \cos(A)} \][/tex]
6. Finally, compare the left-hand side and the right-hand side:
- Left-hand side:
[tex]\[ -\frac{1}{2 \sin(A) \cos(A)} \][/tex]
- Right-hand side:
[tex]\[ \frac{1}{2 \sin(A) \cos(A)} \][/tex]
We find that the simplified form of the left-hand side is:
[tex]\[ -\frac{1}{2 \sin(A) \cos(A)} - \frac{\sin(3A)}{\sin(A)} \cdot \frac{1}{2 \sin(A) \cos(A)} \][/tex]
Since these do not equate ([tex]\((1 + \sin(3A)/\sin(A))\)[/tex] is non-zero), we conclude that:
[tex]\[ -\frac{1}{\tan A} - \frac{1}{\tan 2A} \neq \frac{1}{\sin 2A} \][/tex]
So, the equation simplifies to:
[tex]\[ -(1 + \frac{\sin(3A)}{\sin(A)})/(2 \sin(A) \cos(A)) \neq 0 \][/tex]
This shows that the original equation does not simplify to equality for any general value of [tex]\(A\)[/tex].
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
