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Sagot :
Certainly! Let's go through the problem step-by-step.
Given:
[tex]\[ \tan(x) = \frac{1}{3} \][/tex]
with [tex]\( x \)[/tex] within the interval [tex]\( 90^\circ < x < 270^\circ \)[/tex]. This indicates that [tex]\( x \)[/tex] is in the second or third quadrant. Given that [tex]\( \tan(x) \)[/tex] is positive, [tex]\( x \)[/tex] is in the third quadrant.
We need to find:
[tex]\[ \frac{\sin(2x) - \cos(x)}{2 \tan(x) + \sin(2x)} \][/tex]
### Step-by-Step Solution:
1. Find [tex]\( \sin(x) \)[/tex] and [tex]\( \cos(x) \)[/tex]:
[tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex], thus
[tex]\[ \frac{\sin(x)}{\cos(x)} = \frac{1}{3} \][/tex]
Let:
[tex]\[ \sin(x) = a \][/tex]
[tex]\[ \cos(x) = b \][/tex]
Such that:
[tex]\[ \frac{a}{b} = \frac{1}{3} \][/tex]
[tex]\[ a = \frac{b}{3} \][/tex]
Use the Pythagorean identity:
[tex]\[ \sin^2(x) + \cos^2(x) = 1 \][/tex]
[tex]\[ \left(\frac{b}{3}\right)^2 + b^2 = 1 \][/tex]
[tex]\[ \frac{b^2}{9} + b^2 = 1 \][/tex]
[tex]\[ \frac{b^2}{9} + \frac{9b^2}{9} = 1 \][/tex]
[tex]\[ \frac{10b^2}{9} = 1 \][/tex]
[tex]\[ 10b^2 = 9 \][/tex]
[tex]\[ b^2 = \frac{9}{10} \][/tex]
[tex]\[ b = \pm \frac{3}{\sqrt{10}} \][/tex]
Since [tex]\( x \)[/tex] is in the third quadrant, [tex]\( \cos(x) \)[/tex] is negative:
[tex]\[ \cos(x) = -\frac{3}{\sqrt{10}} \][/tex]
Then,
[tex]\[ \sin(x) = \frac{b}{3} = \frac{-\frac{3}{\sqrt{10}}}{3} = -\frac{1}{\sqrt{10}} \][/tex]
2. Calculate [tex]\( \sin(2x) \)[/tex] and [tex]\( \cos(2x) \)[/tex]:
Using double-angle identities:
[tex]\[ \sin(2x) = 2 \sin(x) \cos(x) \][/tex]
[tex]\[ \sin(2x) = 2 \left(-\frac{1}{\sqrt{10}}\right) \left(-\frac{3}{\sqrt{10}}\right) \][/tex]
[tex]\[ \sin(2x) = 2 \left(\frac{3}{10}\right) \][/tex]
[tex]\[ \sin(2x) = \frac{6}{10} = 0.6 \][/tex]
[tex]\[ \cos(2x) = \cos^2(x) - \sin^2(x) \][/tex]
[tex]\[ \cos(2x) = \left(-\frac{3}{\sqrt{10}}\right)^2 - \left(-\frac{1}{\sqrt{10}}\right)^2 \][/tex]
[tex]\[ \cos(2x) = \frac{9}{10} - \frac{1}{10} \][/tex]
[tex]\[ \cos(2x) = \frac{8}{10} = 0.8 \][/tex]
3. Evaluate the given expression [tex]\(\frac{\sin(2x) - \cos(x)}{2 \tan(x) + \sin(2x)}\)[/tex]:
Numerator:
[tex]\[ \sin(2x) - \cos(x) \][/tex]
[tex]\[ 0.6 - \left(-\frac{3}{\sqrt{10}}\right) \][/tex]
[tex]\[ 0.6 + \frac{3}{\sqrt{10}} \][/tex]
Denominator:
[tex]\[ 2 \tan(x) + \sin(2x) \][/tex]
[tex]\[ 2 \left(\frac{1}{3}\right) + 0.6 \][/tex]
[tex]\[ \frac{2}{3} + 0.6 \][/tex]
Converting 0.6 to a fraction:
[tex]\[ 0.6 = \frac{6}{10} = \frac{3}{5} \][/tex]
Then:
[tex]\[ \frac{2}{3} + \frac{3}{5} \][/tex]
[tex]\[ = \frac{10}{15} + \frac{9}{15} \][/tex]
[tex]\[ = \frac{19}{15} \][/tex]
So, the expression simplifies to:
[tex]\[ \frac{0.6 + \frac{3}{\sqrt{10}}}{\frac{19}{15}} \][/tex]
Simplify numerator:
[tex]\[ 0.6 + \frac{3}{\sqrt{10}} = \frac{6}{10} + \frac{3}{\sqrt{10}} = \frac{1}{\sqrt{10}} \left( \sin(2x) - \cos(x) \right)\][/tex]
Finally:
[tex]\[ \boxed{1.2226447089872476}\][/tex]
Given:
[tex]\[ \tan(x) = \frac{1}{3} \][/tex]
with [tex]\( x \)[/tex] within the interval [tex]\( 90^\circ < x < 270^\circ \)[/tex]. This indicates that [tex]\( x \)[/tex] is in the second or third quadrant. Given that [tex]\( \tan(x) \)[/tex] is positive, [tex]\( x \)[/tex] is in the third quadrant.
We need to find:
[tex]\[ \frac{\sin(2x) - \cos(x)}{2 \tan(x) + \sin(2x)} \][/tex]
### Step-by-Step Solution:
1. Find [tex]\( \sin(x) \)[/tex] and [tex]\( \cos(x) \)[/tex]:
[tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex], thus
[tex]\[ \frac{\sin(x)}{\cos(x)} = \frac{1}{3} \][/tex]
Let:
[tex]\[ \sin(x) = a \][/tex]
[tex]\[ \cos(x) = b \][/tex]
Such that:
[tex]\[ \frac{a}{b} = \frac{1}{3} \][/tex]
[tex]\[ a = \frac{b}{3} \][/tex]
Use the Pythagorean identity:
[tex]\[ \sin^2(x) + \cos^2(x) = 1 \][/tex]
[tex]\[ \left(\frac{b}{3}\right)^2 + b^2 = 1 \][/tex]
[tex]\[ \frac{b^2}{9} + b^2 = 1 \][/tex]
[tex]\[ \frac{b^2}{9} + \frac{9b^2}{9} = 1 \][/tex]
[tex]\[ \frac{10b^2}{9} = 1 \][/tex]
[tex]\[ 10b^2 = 9 \][/tex]
[tex]\[ b^2 = \frac{9}{10} \][/tex]
[tex]\[ b = \pm \frac{3}{\sqrt{10}} \][/tex]
Since [tex]\( x \)[/tex] is in the third quadrant, [tex]\( \cos(x) \)[/tex] is negative:
[tex]\[ \cos(x) = -\frac{3}{\sqrt{10}} \][/tex]
Then,
[tex]\[ \sin(x) = \frac{b}{3} = \frac{-\frac{3}{\sqrt{10}}}{3} = -\frac{1}{\sqrt{10}} \][/tex]
2. Calculate [tex]\( \sin(2x) \)[/tex] and [tex]\( \cos(2x) \)[/tex]:
Using double-angle identities:
[tex]\[ \sin(2x) = 2 \sin(x) \cos(x) \][/tex]
[tex]\[ \sin(2x) = 2 \left(-\frac{1}{\sqrt{10}}\right) \left(-\frac{3}{\sqrt{10}}\right) \][/tex]
[tex]\[ \sin(2x) = 2 \left(\frac{3}{10}\right) \][/tex]
[tex]\[ \sin(2x) = \frac{6}{10} = 0.6 \][/tex]
[tex]\[ \cos(2x) = \cos^2(x) - \sin^2(x) \][/tex]
[tex]\[ \cos(2x) = \left(-\frac{3}{\sqrt{10}}\right)^2 - \left(-\frac{1}{\sqrt{10}}\right)^2 \][/tex]
[tex]\[ \cos(2x) = \frac{9}{10} - \frac{1}{10} \][/tex]
[tex]\[ \cos(2x) = \frac{8}{10} = 0.8 \][/tex]
3. Evaluate the given expression [tex]\(\frac{\sin(2x) - \cos(x)}{2 \tan(x) + \sin(2x)}\)[/tex]:
Numerator:
[tex]\[ \sin(2x) - \cos(x) \][/tex]
[tex]\[ 0.6 - \left(-\frac{3}{\sqrt{10}}\right) \][/tex]
[tex]\[ 0.6 + \frac{3}{\sqrt{10}} \][/tex]
Denominator:
[tex]\[ 2 \tan(x) + \sin(2x) \][/tex]
[tex]\[ 2 \left(\frac{1}{3}\right) + 0.6 \][/tex]
[tex]\[ \frac{2}{3} + 0.6 \][/tex]
Converting 0.6 to a fraction:
[tex]\[ 0.6 = \frac{6}{10} = \frac{3}{5} \][/tex]
Then:
[tex]\[ \frac{2}{3} + \frac{3}{5} \][/tex]
[tex]\[ = \frac{10}{15} + \frac{9}{15} \][/tex]
[tex]\[ = \frac{19}{15} \][/tex]
So, the expression simplifies to:
[tex]\[ \frac{0.6 + \frac{3}{\sqrt{10}}}{\frac{19}{15}} \][/tex]
Simplify numerator:
[tex]\[ 0.6 + \frac{3}{\sqrt{10}} = \frac{6}{10} + \frac{3}{\sqrt{10}} = \frac{1}{\sqrt{10}} \left( \sin(2x) - \cos(x) \right)\][/tex]
Finally:
[tex]\[ \boxed{1.2226447089872476}\][/tex]
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