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Sagot :
Sure, let's go through the process step-by-step to find the value of [tex]\(\sin(\theta)\)[/tex].
Given:
[tex]\[ \tan(\theta) = \frac{5 \sqrt{11}}{11} \][/tex]
### Step 1: Understand the Relationship Between [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex]
Recall the definition of tangent:
[tex]\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \][/tex]
Given [tex]\(\theta\)[/tex] is in quadrant III where both sine and cosine are negative, we can represent:
[tex]\[ \tan(\theta) = \frac{-a}{-b} = \frac{a}{b} \][/tex]
where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are positive.
### Step 2: Calculate [tex]\(\tan^2\theta\)[/tex]
First, find [tex]\(\tan^2(\theta)\)[/tex]:
[tex]\[ \tan^2(\theta) = \left(\frac{5 \sqrt{11}}{11}\right)^2 = \frac{25 \cdot 11}{121} = \frac{275}{121} \approx 2.2727272727272725 \][/tex]
### Step 3: Use the Pythagorean Identity
We use the identity:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
We represent [tex]\(\sin(\theta)\)[/tex] as [tex]\(-a\)[/tex] and [tex]\(\cos(\theta)\)[/tex] as [tex]\(-b\)[/tex], thus:
[tex]\[ (-a)^2 + (-b)^2 = 1 \implies a^2 + b^2 = 1 \][/tex]
### Step 4: Solve for [tex]\(\cos(\theta)\)[/tex] in Terms of [tex]\(\tan(\theta)\)[/tex]
Since [tex]\(\tan(\theta) = \frac{a}{b}\)[/tex]:
[tex]\[ a = \tan(\theta) \cdot b \][/tex]
Substitute [tex]\(a = \tan(\theta) \cdot b\)[/tex] into [tex]\(a^2 + b^2 = 1\)[/tex]:
[tex]\[ (\tan(\theta) \cdot b)^2 + b^2 = 1 \][/tex]
[tex]\[ \tan^2(\theta) \cdot b^2 + b^2 = 1 \][/tex]
[tex]\[ b^2 (\tan^2(\theta) + 1) = 1 \][/tex]
### Step 5: Solve for [tex]\(\cos(\theta)\)[/tex]
[tex]\[ b^2 = \frac{1}{\tan^2(\theta) + 1} \][/tex]
Given [tex]\( \tan^2(\theta) \approx 2.2727272727272725 \)[/tex]:
[tex]\[ b^2 = \frac{1}{2.2727272727272725 + 1} = \frac{1}{3.2727272727272725} \approx 0.3055555555555556 \][/tex]
[tex]\[ b \approx \sqrt{0.3055555555555556} \approx 0.5527707983925667 \][/tex]
### Step 6: Solve for [tex]\(\sin(\theta)\)[/tex]
[tex]\[ a = \tan(\theta) \cdot b \approx 1.507556722888818 \cdot 0.5527707983925667 \approx 0.8333333333333334 \][/tex]
Therefore, since [tex]\(\sin(\theta) = -a\)[/tex] in quadrant III:
[tex]\[ \sin(\theta) \approx -0.8333333333333334 \][/tex]
Thus, the value of [tex]\(\sin(\theta)\)[/tex] is closer to [tex]\( -\frac{5}{6} \)[/tex], making the correct answer:
[tex]\[ \boxed{-\frac{5}{6}} \][/tex]
Given:
[tex]\[ \tan(\theta) = \frac{5 \sqrt{11}}{11} \][/tex]
### Step 1: Understand the Relationship Between [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex]
Recall the definition of tangent:
[tex]\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \][/tex]
Given [tex]\(\theta\)[/tex] is in quadrant III where both sine and cosine are negative, we can represent:
[tex]\[ \tan(\theta) = \frac{-a}{-b} = \frac{a}{b} \][/tex]
where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are positive.
### Step 2: Calculate [tex]\(\tan^2\theta\)[/tex]
First, find [tex]\(\tan^2(\theta)\)[/tex]:
[tex]\[ \tan^2(\theta) = \left(\frac{5 \sqrt{11}}{11}\right)^2 = \frac{25 \cdot 11}{121} = \frac{275}{121} \approx 2.2727272727272725 \][/tex]
### Step 3: Use the Pythagorean Identity
We use the identity:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
We represent [tex]\(\sin(\theta)\)[/tex] as [tex]\(-a\)[/tex] and [tex]\(\cos(\theta)\)[/tex] as [tex]\(-b\)[/tex], thus:
[tex]\[ (-a)^2 + (-b)^2 = 1 \implies a^2 + b^2 = 1 \][/tex]
### Step 4: Solve for [tex]\(\cos(\theta)\)[/tex] in Terms of [tex]\(\tan(\theta)\)[/tex]
Since [tex]\(\tan(\theta) = \frac{a}{b}\)[/tex]:
[tex]\[ a = \tan(\theta) \cdot b \][/tex]
Substitute [tex]\(a = \tan(\theta) \cdot b\)[/tex] into [tex]\(a^2 + b^2 = 1\)[/tex]:
[tex]\[ (\tan(\theta) \cdot b)^2 + b^2 = 1 \][/tex]
[tex]\[ \tan^2(\theta) \cdot b^2 + b^2 = 1 \][/tex]
[tex]\[ b^2 (\tan^2(\theta) + 1) = 1 \][/tex]
### Step 5: Solve for [tex]\(\cos(\theta)\)[/tex]
[tex]\[ b^2 = \frac{1}{\tan^2(\theta) + 1} \][/tex]
Given [tex]\( \tan^2(\theta) \approx 2.2727272727272725 \)[/tex]:
[tex]\[ b^2 = \frac{1}{2.2727272727272725 + 1} = \frac{1}{3.2727272727272725} \approx 0.3055555555555556 \][/tex]
[tex]\[ b \approx \sqrt{0.3055555555555556} \approx 0.5527707983925667 \][/tex]
### Step 6: Solve for [tex]\(\sin(\theta)\)[/tex]
[tex]\[ a = \tan(\theta) \cdot b \approx 1.507556722888818 \cdot 0.5527707983925667 \approx 0.8333333333333334 \][/tex]
Therefore, since [tex]\(\sin(\theta) = -a\)[/tex] in quadrant III:
[tex]\[ \sin(\theta) \approx -0.8333333333333334 \][/tex]
Thus, the value of [tex]\(\sin(\theta)\)[/tex] is closer to [tex]\( -\frac{5}{6} \)[/tex], making the correct answer:
[tex]\[ \boxed{-\frac{5}{6}} \][/tex]
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