Math Problem Statement

Compute tan(pi/12)

Solution

The exact value of tan(π12)\tan\left(\frac{\pi}{12}\right) can be calculated using the identity:

tan(π12)=tan(15)\tan\left(\frac{\pi}{12}\right) = \tan\left(15^\circ\right)

We can express this as the difference of angles formula for tangent:

tan(15)=tan(4530)\tan\left(15^\circ\right) = \tan\left(45^\circ - 30^\circ\right)

Using the tangent subtraction formula:

tan(AB)=tan(A)tan(B)1+tan(A)tan(B)\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}

Substituting A=45A = 45^\circ and B=30B = 30^\circ:

tan(15)=tan(45)tan(30)1+tan(45)tan(30)\tan(15^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ)\tan(30^\circ)}

We know:

tan(45)=1andtan(30)=13\tan(45^\circ) = 1 \quad \text{and} \quad \tan(30^\circ) = \frac{1}{\sqrt{3}}

Now substitute these values:

tan(15)=1131+113\tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}

Simplify the expression:

tan(15)=3133+13\tan(15^\circ) = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}

This simplifies to:

tan(15)=313+1\tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

Thus, the exact value of tan(π12)\tan\left(\frac{\pi}{12}\right) is:

tan(π12)=313+1\tan\left(\frac{\pi}{12}\right) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

Would you like further clarification on this calculation?

Here are 5 related questions:

  1. How can the tangent addition formula be used for other angle combinations?
  2. What is the relationship between radians and degrees in trigonometry?
  3. How can you find the exact values of sine and cosine for common angles?
  4. What are some methods for simplifying trigonometric expressions involving square roots?
  5. How can trigonometric identities be applied to solve real-world problems?

Tip: Memorizing key angle values in both degrees and radians simplifies working with trigonometric identities.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Angle Identities

Formulas

tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A) * tan(B))
tan(45°) = 1
tan(30°) = 1 / sqrt(3)

Theorems

Tangent Subtraction Formula

Suitable Grade Level

Grades 10-12