Math Problem Statement

What if I get: sec(-pi) how would I solve using add 2pi or subtract 2pi

Solution

To solve sec(π)\sec(-\pi), you can use the periodicity of the secant function (a reciprocal of cosine) to simplify the angle by adding or subtracting 2π2\pi. Here’s how you approach it:

Step-by-step solution:

  1. Recall the periodicity of the secant function:

    • The secant function is periodic with a period of 2π2\pi, meaning sec(x+2π)=sec(x)\sec(x + 2\pi) = \sec(x).
  2. Simplify π-\pi using 2π2\pi: To simplify π-\pi within the standard interval [0,2π)[0, 2\pi), you can add 2π2\pi to it: π+2π=π-\pi + 2\pi = \pi So, sec(π)=sec(π)\sec(-\pi) = \sec(\pi).

  3. Evaluate sec(π)\sec(\pi): Recall that sec(x)=1cos(x)\sec(x) = \frac{1}{\cos(x)}. From the unit circle:

    • cos(π)=1\cos(\pi) = -1,
    • Hence, sec(π)=11=1\sec(\pi) = \frac{1}{-1} = -1.

Final Answer:

sec(π)=1\sec(-\pi) = -1


Key Notes:

  • Adding or subtracting 2π2\pi simplifies the angle without changing the value of the trigonometric function.
  • The result shows that sec(π)=sec(π)=1\sec(-\pi) = \sec(\pi) = -1.

Would you like me to provide further examples or details on using 2π2\pi for simplifications?

Related Questions:

  1. What is the period of other trigonometric functions like sine, cosine, and tangent?
  2. How does secant relate to cosine, and what are its key properties?
  3. Can we use symmetry instead of 2π2\pi to solve sec(π)\sec(-\pi)?
  4. How do we evaluate secant for other common angles like π2\frac{\pi}{2} or 3π3\pi?
  5. What happens if you’re given secant for negative angles in degrees instead of radians?

Tip:

Always reduce angles to a principal range (like [0,2π)[0, 2\pi) for radians) to simplify trigonometric evaluations!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Periodicity of Trigonometric Functions

Formulas

\( \sec(x) = \frac{1}{\cos(x)} \)
Periodicity: \( \sec(x + 2\pi) = \sec(x) \)

Theorems

Periodic properties of trigonometric functions
Reciprocal relationship between secant and cosine

Suitable Grade Level

Grades 9-12