To find the exact value of $\tan\left(\frac{7\pi}{8}\right)$ using the half-angle formulas, follow these steps:

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Step 1: Recall the Half-Angle Formula for Tangent

The half-angle formula for tangent is given as: $\tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$ The sign ($+$ or $-$) depends on the quadrant in which $\frac{\theta}{2}$ lies.

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Step 2: Identify $\theta$

In this case, $\frac{7\pi}{8}$ is already a half-angle. Hence, let $\theta = \frac{7\pi}{4}$, so: $\frac{\theta}{2} = \frac{7\pi}{8}$

$\frac{7\pi}{8}$ lies in the second quadrant, where tangent is negative. Thus, we take the negative root of the formula.

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Step 3: Find $\cos\theta$

For $\theta = \frac{7\pi}{4}$, recall the cosine value in the fourth quadrant: $\cos\left(\frac{7\pi}{4}\right) = \cos\left(2\pi - \frac{\pi}{4}\right) = \cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.$

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Step 4: Apply the Half-Angle Formula

Substitute $\cos\theta = \frac{\sqrt{2}}{2}$ into the formula: $\tan\left(\frac{\theta}{2}\right) = -\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}.$ $\tan\left(\frac{7\pi}{8}\right) = -\sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}}.$

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Step 5: Simplify the Numerator and Denominator

1. Simplify the numerator $1 - \frac{\sqrt{2}}{2}$: $1 - \frac{\sqrt{2}}{2} = \frac{2}{2} - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}.$

2. Simplify the denominator $1 + \frac{\sqrt{2}}{2}$: $1 + \frac{\sqrt{2}}{2} = \frac{2}{2} + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}.$

3. Substitute back: $\tan\left(\frac{7\pi}{8}\right) = -\sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}}}.$

4. Simplify the fraction: $\frac{\frac{2 - \sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{2 + \sqrt{2}}.$

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Step 6: Rationalize the Denominator

Multiply numerator and denominator by the conjugate of $2 + \sqrt{2}$, which is $2 - \sqrt{2}$: $\frac{2 - \sqrt{2}}{2 + \sqrt{2}} \cdot \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{(2 - \sqrt{2})^2}{(2 + \sqrt{2})(2 - \sqrt{2})}.$

1. Expand the numerator: $(2 - \sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2}.$

2. Simplify the denominator: $(2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2.$

3. Combine results: $\frac{2 - \sqrt{2}}{2 + \sqrt{2}} = \frac{6 - 4\sqrt{2}}{2} = 3 - 2\sqrt{2}.$

4. Add the negative sign: $\tan\left(\frac{7\pi}{8}\right) = -(3 - 2\sqrt{2}).$

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Final Answer

$\tan\left(\frac{7\pi}{8}\right) = -3 + 2\sqrt{2}.$

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Do you have any questions or would you like further clarification? 😊

Related Questions:

1. How do you derive the half-angle formulas for trigonometric functions?
2. What are the quadrants where tangent is positive or negative?
3. Can the double-angle formula for tangent also solve this problem?
4. How do you calculate exact values of trigonometric functions for special angles?
5. What is the relation between tangent, sine, and cosine in such problems?

Tip:

Always consider the quadrant to determine the correct sign when using half-angle or other trigonometric formulas!