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Find the exact value without a calculator.

[tex]\[
\tan \frac{7 \pi}{12} = -\sqrt{\frac{? + \sqrt{\square}}{-\sqrt{\square}}}
\][/tex]

Double-Angle Formulas:
[tex]\[
\sin(2 \theta) = 2 \sin \theta \cos \theta
\][/tex]
[tex]\[
\cos(2 \theta) = \cos^2 \theta - \sin^2 \theta
\][/tex]
[tex]\[
\tan(2 \theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}
\][/tex]

Half-Angle Formulas:
[tex]\[
\sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}
\][/tex]
[tex]\[
\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}
\][/tex]
[tex]\[
\tan \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}
\][/tex]

Sagot :

GinnyAnswer
To find the exact value of [tex]\(\tan\left(\frac{7\pi}{12}\right)\)[/tex] using a trigonometric approach, we'll follow these steps.

Let's break down the problem systematically.

Step 1: Express [tex]\(\frac{7\pi}{12}\)[/tex] in terms of familiar angles.

[tex]\[ \frac{7\pi}{12} = \frac{3\pi}{4} - \frac{\pi}{3} \][/tex]

So, we need to find [tex]\(\tan\left(\frac{3\pi}{4} - \frac{\pi}{3}\right)\)[/tex].

Step 2: Use the tangent subtraction formula.

The formula for the tangent of the difference of two angles is:

[tex]\[ \tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} \][/tex]

Substituting [tex]\(a = \frac{3\pi}{4}\)[/tex] and [tex]\(b = \frac{\pi}{3}\)[/tex]:

[tex]\[ \tan\left(\frac{3\pi}{4}\right) = -1 \quad \text{(because \(\tan(135^\circ) = -1\))} \][/tex]
[tex]\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \quad \text{(because \(\tan(60^\circ) = \sqrt{3}\))} \][/tex]

Step 3: Substitute the known values into the tangent subtraction formula.

[tex]\[ \tan\left(\frac{3\pi}{4} - \frac{\pi}{3}\right) = \frac{\tan\left(\frac{3\pi}{4}\right) - \tan\left(\frac{\pi}{3}\right)}{1 + \tan\left(\frac{3\pi}{4}\right)\tan\left(\frac{\pi}{3}\right)} = \frac{-1 - \sqrt{3}}{1 + (-1)(\sqrt{3})} = \frac{-1 - \sqrt{3}}{1 - \sqrt{3}} \][/tex]

Step 4: Simplify the expression.

To simplify [tex]\(\frac{-1 - \sqrt{3}}{1 - \sqrt{3}}\)[/tex], multiply the numerator and the denominator by the conjugate of the denominator:

[tex]\[ \frac{-1 - \sqrt{3}}{1 - \sqrt{3}} \cdot \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{(-1 - \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \][/tex]

Calculate the denominator:

[tex]\[ (1 - \sqrt{3})(1 + \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2 \][/tex]

Calculate the numerator:

[tex]\[ (-1 - \sqrt{3})(1 + \sqrt{3}) = (-1)(1) + (-1)(\sqrt{3}) + (-\sqrt{3})(1) + (-\sqrt{3})(\sqrt{3}) = -1 - \sqrt{3} - \sqrt{3} - 3 = -4 - 2\sqrt{3} \][/tex]

So the expression becomes:

[tex]\[ \frac{-1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{-4 - 2\sqrt{3}}{-2} = 2 + \sqrt{3} \][/tex]

Rewriting it for the final exact form:

[tex]\[ \tan\left(\frac{7\pi}{12}\right) = -\sqrt{\frac{-4 + 2\sqrt{3}}{-2}} = 2 + \sqrt{3} \][/tex]

Hence the identity for [tex]\(\tan\left(\frac{7\pi}{12}\right) = 2 + \sqrt{3}\)[/tex]
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