IDNLearn.com: Your destination for reliable and timely answers to any question. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.
Sagot :
To find the value of [tex]\(\tan \theta\)[/tex] given that [tex]\(\sin \theta = \frac{2 \sqrt{x}}{x + 1}\)[/tex], we can follow these steps:
1. Compute [tex]\(\sin \theta\)[/tex]:
We are given:
[tex]\[ \sin \theta = \frac{2 \sqrt{x}}{x + 1} \][/tex]
2. Calculate [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta = \left(\frac{2 \sqrt{x}}{x + 1}\right)^2 = \frac{4x}{(x + 1)^2} \][/tex]
3. Use the Pythagorean identity to find [tex]\(\cos^2 \theta\)[/tex]:
We know that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Therefore:
[tex]\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{4x}{(x + 1)^2} \][/tex]
4. Simplify [tex]\(\cos^2 \theta\)[/tex]:
Finding a common denominator:
[tex]\[ \cos^2 \theta = \frac{(x + 1)^2 - 4x}{(x + 1)^2} \][/tex]
Simplify the numerator:
[tex]\[ (x + 1)^2 - 4x = x^2 + 2x + 1 - 4x = x^2 - 2x + 1 = (x - 1)^2 \][/tex]
Therefore:
[tex]\[ \cos^2 \theta = \frac{(x - 1)^2}{(x + 1)^2} \][/tex]
5. Find [tex]\(\cos \theta\)[/tex]:
Taking the positive square root (assuming [tex]\(\cos \theta\)[/tex] is positive):
[tex]\[ \cos \theta = \frac{|x - 1|}{x + 1} \][/tex]
6. Calculate [tex]\(\tan \theta\)[/tex]:
Using the definition of [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2 \sqrt{x}}{x + 1}}{\frac{|x - 1|}{x + 1}} = \frac{2 \sqrt{x}}{|x - 1|} \][/tex]
Now, let's consider an example to verify the calculations with a specific value for [tex]\(x\)[/tex]. Let [tex]\(x = 4\)[/tex]:
- [tex]\(\sin \theta = \frac{2 \sqrt{4}}{4 + 1} = \frac{2 \cdot 2}{5} = \frac{4}{5} = 0.8\)[/tex]
- [tex]\(\cos \theta = \frac{|4 - 1|}{4 + 1} = \frac{3}{5} = 0.6\)[/tex]
- [tex]\(\tan \theta = \frac{0.8}{0.6} \approx 1.3333\)[/tex]
Therefore, for [tex]\(x = 4\)[/tex], we have:
- [tex]\(\sin \theta = 0.8\)[/tex]
- [tex]\(\cos \theta = 0.6\)[/tex]
- [tex]\(\tan \theta \approx 1.3333\)[/tex]
Thus, the value of [tex]\(\tan \theta\)[/tex] is approximately 1.3333 when [tex]\(x = 4\)[/tex].
1. Compute [tex]\(\sin \theta\)[/tex]:
We are given:
[tex]\[ \sin \theta = \frac{2 \sqrt{x}}{x + 1} \][/tex]
2. Calculate [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta = \left(\frac{2 \sqrt{x}}{x + 1}\right)^2 = \frac{4x}{(x + 1)^2} \][/tex]
3. Use the Pythagorean identity to find [tex]\(\cos^2 \theta\)[/tex]:
We know that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Therefore:
[tex]\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{4x}{(x + 1)^2} \][/tex]
4. Simplify [tex]\(\cos^2 \theta\)[/tex]:
Finding a common denominator:
[tex]\[ \cos^2 \theta = \frac{(x + 1)^2 - 4x}{(x + 1)^2} \][/tex]
Simplify the numerator:
[tex]\[ (x + 1)^2 - 4x = x^2 + 2x + 1 - 4x = x^2 - 2x + 1 = (x - 1)^2 \][/tex]
Therefore:
[tex]\[ \cos^2 \theta = \frac{(x - 1)^2}{(x + 1)^2} \][/tex]
5. Find [tex]\(\cos \theta\)[/tex]:
Taking the positive square root (assuming [tex]\(\cos \theta\)[/tex] is positive):
[tex]\[ \cos \theta = \frac{|x - 1|}{x + 1} \][/tex]
6. Calculate [tex]\(\tan \theta\)[/tex]:
Using the definition of [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2 \sqrt{x}}{x + 1}}{\frac{|x - 1|}{x + 1}} = \frac{2 \sqrt{x}}{|x - 1|} \][/tex]
Now, let's consider an example to verify the calculations with a specific value for [tex]\(x\)[/tex]. Let [tex]\(x = 4\)[/tex]:
- [tex]\(\sin \theta = \frac{2 \sqrt{4}}{4 + 1} = \frac{2 \cdot 2}{5} = \frac{4}{5} = 0.8\)[/tex]
- [tex]\(\cos \theta = \frac{|4 - 1|}{4 + 1} = \frac{3}{5} = 0.6\)[/tex]
- [tex]\(\tan \theta = \frac{0.8}{0.6} \approx 1.3333\)[/tex]
Therefore, for [tex]\(x = 4\)[/tex], we have:
- [tex]\(\sin \theta = 0.8\)[/tex]
- [tex]\(\cos \theta = 0.6\)[/tex]
- [tex]\(\tan \theta \approx 1.3333\)[/tex]
Thus, the value of [tex]\(\tan \theta\)[/tex] is approximately 1.3333 when [tex]\(x = 4\)[/tex].
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.