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Sagot :
To find the value of [tex]\(\tan \frac{7 \pi}{12}\)[/tex], let's break down the problem using trigonometric identities, specifically the angle sum identity for tangent. The angle sum identity for tangent is given by:
[tex]\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \][/tex]
We need to express [tex]\(\frac{7 \pi}{12}\)[/tex] in a form that is more manageable. Notice that:
[tex]\[ \frac{7 \pi}{12} = \frac{3 \pi}{12} + \frac{4 \pi}{12} = \frac{\pi}{4} + \frac{\pi}{6} \][/tex]
So, set [tex]\( A = \frac{\pi}{4} \)[/tex] and [tex]\( B = \frac{\pi}{6} \)[/tex].
Next, we use the known values:
[tex]\[ \tan \left( \frac{\pi}{4} \right) = 1 \][/tex]
[tex]\[ \tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \][/tex]
Using the angle sum identity:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \tan \left( \frac{\pi}{4} + \frac{\pi}{6} \right) = \frac{\tan \left( \frac{\pi}{4} \right) + \tan \left( \frac{\pi}{6} \right)}{1 - \tan \left( \frac{\pi}{4} \right) \tan \left( \frac{\pi}{6} \right)} \][/tex]
Substitute the values:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} \][/tex]
Simplify the numerator and the denominator separately:
[tex]\[ \text{Numerator: } 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \][/tex]
[tex]\[ \text{Denominator: } 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \][/tex]
Combine these results:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \][/tex]
At this point, we rationalize the result:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \][/tex]
Numeric evaluation of [tex]\(2 + \sqrt{3}\)[/tex]:
[tex]\[ 2 + \sqrt{3} \approx 3.732050807568876 \][/tex]
Thus, the value of [tex]\(\tan \frac{7 \pi}{12}\)[/tex] is:
[tex]\[ \boxed{3.732050807568876} \][/tex]
[tex]\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \][/tex]
We need to express [tex]\(\frac{7 \pi}{12}\)[/tex] in a form that is more manageable. Notice that:
[tex]\[ \frac{7 \pi}{12} = \frac{3 \pi}{12} + \frac{4 \pi}{12} = \frac{\pi}{4} + \frac{\pi}{6} \][/tex]
So, set [tex]\( A = \frac{\pi}{4} \)[/tex] and [tex]\( B = \frac{\pi}{6} \)[/tex].
Next, we use the known values:
[tex]\[ \tan \left( \frac{\pi}{4} \right) = 1 \][/tex]
[tex]\[ \tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \][/tex]
Using the angle sum identity:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \tan \left( \frac{\pi}{4} + \frac{\pi}{6} \right) = \frac{\tan \left( \frac{\pi}{4} \right) + \tan \left( \frac{\pi}{6} \right)}{1 - \tan \left( \frac{\pi}{4} \right) \tan \left( \frac{\pi}{6} \right)} \][/tex]
Substitute the values:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} \][/tex]
Simplify the numerator and the denominator separately:
[tex]\[ \text{Numerator: } 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \][/tex]
[tex]\[ \text{Denominator: } 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \][/tex]
Combine these results:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \][/tex]
At this point, we rationalize the result:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \][/tex]
Numeric evaluation of [tex]\(2 + \sqrt{3}\)[/tex]:
[tex]\[ 2 + \sqrt{3} \approx 3.732050807568876 \][/tex]
Thus, the value of [tex]\(\tan \frac{7 \pi}{12}\)[/tex] is:
[tex]\[ \boxed{3.732050807568876} \][/tex]
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