Proving Trigonometric Equality Using Induction: A Comprehensive Guide

In this article, we will explore the process of proving the trigonometric identity tan(2^n x) cot(2^n x) - 2cot(2^{n-1} x). This proof employs the method of mathematical induction and leverages various trigonometric identities. By the end of this guide, you will have a thorough understanding of how to prove this identity step by step.

1. Introduction to the Identity

The goal is to prove the identity:

tan(2^n x) cot(2^n x) - 2cot(2^{n-1} x)

To start, let's rewrite the trigonometric functions tan, cot, and cot in terms of sine and cosine. Recall the following definitions:

tan(2^n x) frac{sin(2^n x)}{cos(2^n x)}

cot(2^n x) frac{cos(2^n x)}{sin(2^n x)}

cot(2^{n-1} x) frac{cos(2^{n-1} x)}{sin(2^{n-1} x)}

We can now proceed with the proof using the method of mathematical induction.

2. Base Case

First, we will consider the base case, where n 0:

cot(2^{0-1} x) frac{cot(2^0 x) - tan(2^0 x)}{2}

This simplifies to:

cot(2x) frac{cot(x) - tan(x)}{2}

To show that this holds, we use the cotangent of the double angle identity:

cot(2x) frac{cot^2(x) - 1}{2cot(x)}

Simplifying the right-hand side, we get:

cot(2x) frac{cot^2(x)}{2cot(x)} - frac{1}{2cot(x)}

cot(2x) frac{cot(x)}{2} - frac{tan(x)}{2}

cot(2x) frac{cot(x) - tan(x)}{2}

Hence, the base case holds true.

3. Inductive Hypothesis

For an arbitrary integer k, assume that the identity holds:

cot(2^{k-1} x) frac{cot(2^k x) - tan(2^k x)}{2}

4. Inductive Step

We aim to show that the identity holds for k 1:

cot(2^{(k 1)-1} x) frac{cot(2^{k 1} x) - tan(2^{k 1} x)}{2}

cot(2^{k 1-1} x) frac{cot(2^k x) - tan(2^k x)}{2}

Note that:

cot(2^{(k 1)-1} x) cot(2 cdot 2^{k-1} x) cot(2 theta), where theta 2^{k-1} x.

Using the cotangent of the double angle identity, we get:

cot(2theta) frac{cot^2(theta) - 1}{2cot(theta)}

Substituting theta 2^{k-1} x, we have:

cot(2^{k 1-1} x) frac{cot^2(2^{k-1} x) - 1}{2cot(2^{k-1} x)}

cot(2^{k 1-1} x) frac{cot(2^{k-1} x)}{2} - frac{1}{2cot(2^{k-1} x)}

cot(2^{k 1-1} x) frac{cot(2^{k-1} x) - tan(2^{k-1} x)}{2}

Thus, the inductive step holds.

Since the base case and the inductive step are both true, we conclude that:

cot(2^{n-1} x) frac{cot(2^n x) - tan(2^n x)}{2}

And by rearranging, we get:

tan(2^n x) cot(2^n x) - 2cot(2^{n-1} x)

Hence, the identity is proven.

5. Conclusion

The proof involves the use of trigonometric identities and mathematical induction. The equality holds true under these transformations, thus verifying the original statement. Understanding and practicing such proofs can greatly enhance your knowledge of trigonometric functions and their properties.