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Sagot :
To find the exact value of [tex]\(\tan \frac{7\pi}{12}\)[/tex], we need to use the half-angle formula for tangent. Specifically, we will use the formula:
[tex]\[ \tan \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \][/tex]
First, we express [tex]\(\frac{7\pi}{12}\)[/tex] as a half-angle. Notice that:
[tex]\[ \frac{7\pi}{12} = \frac{7\pi}{6} \times \frac{1}{2} \][/tex]
Thus, [tex]\(\theta = \frac{7\pi}{6}\)[/tex], which means we are dealing with:
[tex]\[ \tan \left( \frac{\frac{7\pi}{6}}{2} \right) = \tan \left( \frac{7\pi}{12} \right) \][/tex]
Using the [tex]\(\tan\)[/tex] half-angle formula, we have:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{1 - \cos \left( \frac{7\pi}{6} \right)}{1 + \cos \left( \frac{7\pi}{6} \right)}} \][/tex]
Next, we need to find [tex]\(\cos \frac{7\pi}{6}\)[/tex]. Note that [tex]\(\frac{7\pi}{6}\)[/tex] is in the third quadrant where cosine is negative. We know that:
[tex]\[ \cos \left( \frac{7\pi}{6} \right) = \cos \left( \pi + \frac{\pi}{6} \right) = -\cos \left( \frac{\pi}{6} \right) \][/tex]
Since [tex]\(\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}\)[/tex], we have:
[tex]\[ \cos \left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2} \][/tex]
Substitute this back into the half-angle formula:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{1 - \left( -\frac{\sqrt{3}}{2} \right)}{1 + \left( -\frac{\sqrt{3}}{2} \right)}} = \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} \][/tex]
Now, simplify the expression inside the square root:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} = \pm \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}} = \pm \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}} \][/tex]
To rationalize the denominator, multiply by the conjugate of the denominator:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}} = \pm \sqrt{\frac{(2 + \sqrt{3})^2}{4 - 3}} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{4 + 4\sqrt{3} + 3}{1}} = \pm \sqrt{7 + 4\sqrt{3}} \][/tex]
Therefore, we find that:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{7 + 4\sqrt{3}} \][/tex]
In the relevant equation [tex]\(\tan \frac{7\pi}{12} = -\sqrt{\frac{[?] + \sqrt{\square}}{\square - \sqrt{\square}}}\)[/tex], the exact value corresponds to:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = -\sqrt{\frac{7 + 4\sqrt{3}}{1}} \][/tex]
From this, we can match the components:
- The unknown value [tex]\([?]\)[/tex] is [tex]\(7\)[/tex]
- Both occurrences of [tex]\(\sqrt{\square}\)[/tex] are [tex]\(\sqrt{3}\)[/tex]
Thus, the exact value is:
[tex]\[ -\sqrt{\frac{7 + 4\sqrt{3}}{1}} \][/tex]
[tex]\[ \tan \left( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \][/tex]
First, we express [tex]\(\frac{7\pi}{12}\)[/tex] as a half-angle. Notice that:
[tex]\[ \frac{7\pi}{12} = \frac{7\pi}{6} \times \frac{1}{2} \][/tex]
Thus, [tex]\(\theta = \frac{7\pi}{6}\)[/tex], which means we are dealing with:
[tex]\[ \tan \left( \frac{\frac{7\pi}{6}}{2} \right) = \tan \left( \frac{7\pi}{12} \right) \][/tex]
Using the [tex]\(\tan\)[/tex] half-angle formula, we have:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{1 - \cos \left( \frac{7\pi}{6} \right)}{1 + \cos \left( \frac{7\pi}{6} \right)}} \][/tex]
Next, we need to find [tex]\(\cos \frac{7\pi}{6}\)[/tex]. Note that [tex]\(\frac{7\pi}{6}\)[/tex] is in the third quadrant where cosine is negative. We know that:
[tex]\[ \cos \left( \frac{7\pi}{6} \right) = \cos \left( \pi + \frac{\pi}{6} \right) = -\cos \left( \frac{\pi}{6} \right) \][/tex]
Since [tex]\(\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}\)[/tex], we have:
[tex]\[ \cos \left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2} \][/tex]
Substitute this back into the half-angle formula:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{1 - \left( -\frac{\sqrt{3}}{2} \right)}{1 + \left( -\frac{\sqrt{3}}{2} \right)}} = \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} \][/tex]
Now, simplify the expression inside the square root:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} = \pm \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}} = \pm \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}} \][/tex]
To rationalize the denominator, multiply by the conjugate of the denominator:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}} = \pm \sqrt{\frac{(2 + \sqrt{3})^2}{4 - 3}} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{\frac{4 + 4\sqrt{3} + 3}{1}} = \pm \sqrt{7 + 4\sqrt{3}} \][/tex]
Therefore, we find that:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \pm \sqrt{7 + 4\sqrt{3}} \][/tex]
In the relevant equation [tex]\(\tan \frac{7\pi}{12} = -\sqrt{\frac{[?] + \sqrt{\square}}{\square - \sqrt{\square}}}\)[/tex], the exact value corresponds to:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = -\sqrt{\frac{7 + 4\sqrt{3}}{1}} \][/tex]
From this, we can match the components:
- The unknown value [tex]\([?]\)[/tex] is [tex]\(7\)[/tex]
- Both occurrences of [tex]\(\sqrt{\square}\)[/tex] are [tex]\(\sqrt{3}\)[/tex]
Thus, the exact value is:
[tex]\[ -\sqrt{\frac{7 + 4\sqrt{3}}{1}} \][/tex]
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