Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To express [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex], follow these steps:
1. Understand the function [tex]\(\cos^{-1} x\)[/tex]:
- [tex]\(\cos^{-1} x\)[/tex] (or arccosine of [tex]\(x\)[/tex]) is the angle [tex]\(\theta\)[/tex] such that [tex]\(\cos(\theta) = x\)[/tex].
- Thus, denote [tex]\(\theta = \cos^{-1} x\)[/tex]. Consequently, [tex]\(\cos(\theta) = x\)[/tex].
2. Express [tex]\(\cot(\theta)\)[/tex] in terms of trigonometric functions:
- The cotangent function is given by: [tex]\(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\)[/tex].
3. Substitute [tex]\(\cos(\theta)\)[/tex]:
- Since [tex]\(\cos(\theta) = x\)[/tex], we need to find [tex]\(\sin(\theta)\)[/tex].
4. Use the Pythagorean identity for sine:
- The identity [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex] helps here.
- Substitute [tex]\(\cos(\theta) = x\)[/tex] into the identity: [tex]\(\sin^2(\theta) + x^2 = 1\)[/tex].
- Solving for [tex]\(\sin^2(\theta)\)[/tex]: [tex]\(\sin^2(\theta) = 1 - x^2\)[/tex].
- Then, [tex]\(\sin(\theta) = \sqrt{1 - x^2}\)[/tex] (we take the positive root because [tex]\(\theta = \cos^{-1} x\)[/tex] gives an angle in the range [tex]\([0, \pi]\)[/tex], where sine is non-negative).
5. Combine the expressions:
- Now, substitute [tex]\(\cos(\theta) = x\)[/tex] and [tex]\(\sin(\theta) = \sqrt{1 - x^2}\)[/tex] into the cotangent expression:
[tex]\[ \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{x}{\sqrt{1 - x^2}} \][/tex]
Hence, the expression [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex] is:
[tex]\[ \cot \left(\cos^{-1} x\right) = \frac{x}{\sqrt{1 - x^2}} \][/tex]
This completes the step-by-step process of rewriting [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex].
1. Understand the function [tex]\(\cos^{-1} x\)[/tex]:
- [tex]\(\cos^{-1} x\)[/tex] (or arccosine of [tex]\(x\)[/tex]) is the angle [tex]\(\theta\)[/tex] such that [tex]\(\cos(\theta) = x\)[/tex].
- Thus, denote [tex]\(\theta = \cos^{-1} x\)[/tex]. Consequently, [tex]\(\cos(\theta) = x\)[/tex].
2. Express [tex]\(\cot(\theta)\)[/tex] in terms of trigonometric functions:
- The cotangent function is given by: [tex]\(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\)[/tex].
3. Substitute [tex]\(\cos(\theta)\)[/tex]:
- Since [tex]\(\cos(\theta) = x\)[/tex], we need to find [tex]\(\sin(\theta)\)[/tex].
4. Use the Pythagorean identity for sine:
- The identity [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex] helps here.
- Substitute [tex]\(\cos(\theta) = x\)[/tex] into the identity: [tex]\(\sin^2(\theta) + x^2 = 1\)[/tex].
- Solving for [tex]\(\sin^2(\theta)\)[/tex]: [tex]\(\sin^2(\theta) = 1 - x^2\)[/tex].
- Then, [tex]\(\sin(\theta) = \sqrt{1 - x^2}\)[/tex] (we take the positive root because [tex]\(\theta = \cos^{-1} x\)[/tex] gives an angle in the range [tex]\([0, \pi]\)[/tex], where sine is non-negative).
5. Combine the expressions:
- Now, substitute [tex]\(\cos(\theta) = x\)[/tex] and [tex]\(\sin(\theta) = \sqrt{1 - x^2}\)[/tex] into the cotangent expression:
[tex]\[ \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{x}{\sqrt{1 - x^2}} \][/tex]
Hence, the expression [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex] is:
[tex]\[ \cot \left(\cos^{-1} x\right) = \frac{x}{\sqrt{1 - x^2}} \][/tex]
This completes the step-by-step process of rewriting [tex]\(\cot \left(\cos^{-1} x\right)\)[/tex] in terms of [tex]\(x\)[/tex].
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.