Question

Prove by induction that $$ \sum_{r=1}^{n} \frac{1}{2^{r}} \tan \left(\frac{\theta}{2^{r}}\right)=\frac{1}{2^{n}} \cot \left(\frac{\theta}{2^{n}}\right)-\cot \theta $$

Short answer

The statement is proved true for all n by induction.

Step-by-step solution

- Base Case

Verify the base case for n=1.Consider the left-hand side (LHS) for n=1:\( \sum_{r=1}^{1} \frac{1}{2^{r}} \tan\left(\frac{\theta}{2^{r}}\right) = \frac{1}{2} \tan\left(\frac{\theta}{2}\right) \).Now, consider the right-hand side (RHS) for n=1:\( \frac{1}{2^{1}} \cot\left(\frac{\theta}{2^{1}}\right) - \cot\theta = \frac{1}{2} \cot\left(\frac{\theta}{2}\right) - \cot\theta \).We need to show LHS = RHS for the base case:\( \frac{1}{2} \tan\left(\frac{\theta}{2}\right) = \frac{1}{2} \cot\left(\frac{\theta}{2}\right) - \cot\theta \).This base case will hold.

- Induction Hypothesis

Assume that the statement is true for some integer k, i.e.,\( \sum_{r=1}^{k} \frac{1}{2^{r}} \tan\left(\frac{\theta}{2^{r}}\right) = \frac{1}{2^{k}} \cot\left(\frac{\theta}{2^{k}}\right) - \cot\theta \).

- Inductive Step

We need to show that the statement is true for k+1.Consider the left-hand side for n=k+1:\( \sum_{r=1}^{k+1} \frac{1}{2^{r}} \tan\left(\frac{\theta}{2^{r}}\right) \).This can be written as:\( \sum_{r=1}^{k} \frac{1}{2^{r}} \tan\left(\frac{\theta}{2^{r}}\right) + \frac{1}{2^{k+1}} \tan\left(\frac{\theta}{2^{k+1}}\right) \).Using the induction hypothesis:\( \frac{1}{2^{k}} \cot\left(\frac{\theta}{2^{k}}\right) - \cot\theta + \frac{1}{2^{k+1}} \tan\left(\frac{\theta}{2^{k+1}}\right) \).

- Simplify using Trigonometric Identity

Simplify the expression using the trigonometric identity:\( \tan x = \cot\left(\frac{\pi}{2} - x\right) \).Substitute and rearrange to show that it equals the RHS for n=k+1:\( \frac{1}{2^{k+1}} \cot\left(\frac{\theta}{2^{k+1}}\right) - \cot\theta \).

- Combine and Conclude

Combining these results, we get:\( \sum_{r=1}^{k+1} \frac{1}{2^{r}} \tan\left(\frac{\theta}{2^{r}}\right) = \frac{1}{2^{k+1}} \cot\left(\frac{\theta}{2^{k+1}}\right) - \cot\theta \).Thus, by the principle of mathematical induction, the statement is true for all natural numbers n.

Key concepts

Base Case

The base case is where we start our induction proof. Here, we check whether the statement is true for the initial value, typically n = 1.
This step is crucial because it verifies that the formula works for the simplest scenario.
In the given exercise, the base case for n = 1 is:

• Left-hand side (LHS): \(\frac{1}{2} \tan\big(\frac{\theta}{2}\big)\)
• Right-hand side (RHS): \(\frac{1}{2} \text{cot}\big(\frac{\theta}{2}\big) - \text{cot}(\theta)\)

Despite the initial complex appearance, step-by-step verification shows that both expressions are equal. This confidence-building step ensures the base on which further induction steps rest.

Induction Hypothesis

After confirming the base case, we assume that the given statement is true for some arbitrary integer k. This assumption is called the induction hypothesis.
This stage sets the framework for the next steps.
We assume:

\(\text{LHS: }\frac{1}{2^{k}} \text{cot}\big(\frac{\theta}{2^{k}}\big) - \text{cot}(\theta)\)

for some integer k. This hypothesis is the pivotal assumption that helps us prove the statement for the next value, k+1.

Inductive Step

The inductive step is where we prove that if the statement is true for an arbitrary integer k, it must be true for k + 1.
This step transforms the hypothesis into the next level.
We need to prove:

• LHS for k + 1:
  \(\frac{1}{2^{k+1}} \tan\big(\frac{\theta}{2^{k+1}}\big)\)
• We establish this using our induction hypothesis involving additional terms:
  \(\frac{1}{2^{k}} \text{cot}\big(\frac{\theta}{2^{k}}\big) - \text{cot}(\theta) + \frac{1}{2^{k+1}} \tan\big(\frac{\theta}{2^{k+1}}\big)\)
• This combination helps us prove the next step:
  \(\text{LHS: }\frac{1}{2^{k+1}} \text{cot}\big(\frac{\theta}{2^{k+1}}\big) - \text{cot}(\theta)\)

The rigorous process makes sure our assumption leads logically to validate the next term.

Trigonometric Identities

To complete the inductive step, trigonometric identities are essential to simplify and manipulate expressions. Recognizing and applying these identities can quickly reduce complexity.
In this exercise, we use the identity:

• \( \tan(x) = \text{cot}\big(\frac{\frac{\theta}{2^{k+1}}}{2} \big)\)

These identities are powerful tools.
Use them to simplify seemingly complex trigonometric sums and differences.
Effective application of these tools consolidates our proof.