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Sagot :
Let's solve the given problem step by step.
Given the equation [tex]\(5 \tan \beta = 4\)[/tex], we want to determine the value of [tex]\(\frac{5 \sin \beta - 2 \cos \beta}{5 \sin \beta + 2 \cos \beta}\)[/tex].
1. Express [tex]\(\tan \beta\)[/tex] in a simplified form:
[tex]\[ \tan \beta = \frac{4}{5} \][/tex]
Since [tex]\(\tan \beta = \frac{\sin \beta}{\cos \beta}\)[/tex], we can write:
[tex]\[ \frac{\sin \beta}{\cos \beta} = \frac{4}{5} \][/tex]
2. Express [tex]\(\sin \beta\)[/tex] in terms of [tex]\(\cos \beta\)[/tex]:
[tex]\[ \sin \beta = \frac{4}{5} \cos \beta \][/tex]
3. Use the Pythagorean identity [tex]\(\sin^2 \beta + \cos^2 \beta = 1\)[/tex] to find [tex]\(\cos \beta\)[/tex]:
[tex]\[ \left( \frac{4}{5} \cos \beta \right)^2 + \cos^2 \beta = 1 \][/tex]
[tex]\[ \frac{16}{25} \cos^2 \beta + \cos^2 \beta = 1 \][/tex]
[tex]\[ \left( \frac{16}{25} + 1 \right) \cos^2 \beta = 1 \][/tex]
[tex]\[ \frac{41}{25} \cos^2 \beta = 1 \][/tex]
[tex]\[ \cos^2 \beta = \frac{25}{41} \][/tex]
Therefore:
[tex]\[ \cos \beta = \sqrt{ \frac{25}{41} } = \frac{5}{\sqrt{41}} \][/tex]
4. Find [tex]\(\sin \beta\)[/tex]:
[tex]\[ \sin \beta = \frac{4}{5} \cos \beta = \frac{4}{5} \cdot \frac{5}{\sqrt{41}} = \frac{4}{\sqrt{41}} \][/tex]
5. Calculate the numerator [tex]\(5 \sin \beta - 2 \cos \beta\)[/tex]:
[tex]\[ 5 \sin \beta - 2 \cos \beta = 5 \left( \frac{4}{\sqrt{41}} \right) - 2 \left( \frac{5}{\sqrt{41}} \right) \][/tex]
[tex]\[ = \frac{20}{\sqrt{41}} - \frac{10}{\sqrt{41}} \][/tex]
[tex]\[ = \frac{10}{\sqrt{41}} \][/tex]
6. Calculate the denominator [tex]\(5 \sin \beta + 2 \cos \beta\)[/tex]:
[tex]\[ 5 \sin \beta + 2 \cos \beta = 5 \left( \frac{4}{\sqrt{41}} \right) + 2 \left( \frac{5}{\sqrt{41}} \right) \][/tex]
[tex]\[ = \frac{20}{\sqrt{41}} + \frac{10}{\sqrt{41}} \][/tex]
[tex]\[ = \frac{30}{\sqrt{41}} \][/tex]
7. Form the required expression:
[tex]\[ \frac{5 \sin \beta - 2 \cos \beta}{5 \sin \beta + 2 \cos \beta} = \frac{\frac{10}{\sqrt{41}}}{\frac{30}{\sqrt{41}}} = \frac{10}{30} = \frac{1}{3} \][/tex]
Thus, the answer is:
[tex]\[ \boxed{\frac{1}{3}} \][/tex]
Given the equation [tex]\(5 \tan \beta = 4\)[/tex], we want to determine the value of [tex]\(\frac{5 \sin \beta - 2 \cos \beta}{5 \sin \beta + 2 \cos \beta}\)[/tex].
1. Express [tex]\(\tan \beta\)[/tex] in a simplified form:
[tex]\[ \tan \beta = \frac{4}{5} \][/tex]
Since [tex]\(\tan \beta = \frac{\sin \beta}{\cos \beta}\)[/tex], we can write:
[tex]\[ \frac{\sin \beta}{\cos \beta} = \frac{4}{5} \][/tex]
2. Express [tex]\(\sin \beta\)[/tex] in terms of [tex]\(\cos \beta\)[/tex]:
[tex]\[ \sin \beta = \frac{4}{5} \cos \beta \][/tex]
3. Use the Pythagorean identity [tex]\(\sin^2 \beta + \cos^2 \beta = 1\)[/tex] to find [tex]\(\cos \beta\)[/tex]:
[tex]\[ \left( \frac{4}{5} \cos \beta \right)^2 + \cos^2 \beta = 1 \][/tex]
[tex]\[ \frac{16}{25} \cos^2 \beta + \cos^2 \beta = 1 \][/tex]
[tex]\[ \left( \frac{16}{25} + 1 \right) \cos^2 \beta = 1 \][/tex]
[tex]\[ \frac{41}{25} \cos^2 \beta = 1 \][/tex]
[tex]\[ \cos^2 \beta = \frac{25}{41} \][/tex]
Therefore:
[tex]\[ \cos \beta = \sqrt{ \frac{25}{41} } = \frac{5}{\sqrt{41}} \][/tex]
4. Find [tex]\(\sin \beta\)[/tex]:
[tex]\[ \sin \beta = \frac{4}{5} \cos \beta = \frac{4}{5} \cdot \frac{5}{\sqrt{41}} = \frac{4}{\sqrt{41}} \][/tex]
5. Calculate the numerator [tex]\(5 \sin \beta - 2 \cos \beta\)[/tex]:
[tex]\[ 5 \sin \beta - 2 \cos \beta = 5 \left( \frac{4}{\sqrt{41}} \right) - 2 \left( \frac{5}{\sqrt{41}} \right) \][/tex]
[tex]\[ = \frac{20}{\sqrt{41}} - \frac{10}{\sqrt{41}} \][/tex]
[tex]\[ = \frac{10}{\sqrt{41}} \][/tex]
6. Calculate the denominator [tex]\(5 \sin \beta + 2 \cos \beta\)[/tex]:
[tex]\[ 5 \sin \beta + 2 \cos \beta = 5 \left( \frac{4}{\sqrt{41}} \right) + 2 \left( \frac{5}{\sqrt{41}} \right) \][/tex]
[tex]\[ = \frac{20}{\sqrt{41}} + \frac{10}{\sqrt{41}} \][/tex]
[tex]\[ = \frac{30}{\sqrt{41}} \][/tex]
7. Form the required expression:
[tex]\[ \frac{5 \sin \beta - 2 \cos \beta}{5 \sin \beta + 2 \cos \beta} = \frac{\frac{10}{\sqrt{41}}}{\frac{30}{\sqrt{41}}} = \frac{10}{30} = \frac{1}{3} \][/tex]
Thus, the answer is:
[tex]\[ \boxed{\frac{1}{3}} \][/tex]
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