Nested Radicals and a Cosine Formula

What do nested square roots have to do with trigonometry? Here we explore a sequence defined by $a_1 = \sqrt{2}$, $a_{n+1} = \sqrt{2+a_n}$, and prove by induction that $a_n = 2\cos(\pi/2^{n+1})$. This elegant connection between algebraic recurrence and trigonometric closed form is a favourite in HSC Extension 2 proofs, and we'll walk through the induction step carefully, focusing on the double-angle identity and sign analysis.

Problem Statement

The numbers $a_n$, for integers $n \geq 1$, are defined as:

$$\begin{aligned} a_1 &= \sqrt{2} \\ a_2 &= \sqrt{2 + \sqrt{2}} \\ a_3 &= \sqrt{2 + \sqrt{2 + \sqrt{2}}}, \text{ and so on.} \end{aligned}$$

These numbers satisfy the relation $a_{n+1}^2 = 2 + a_n$, for $n \geq 1$. (Do NOT prove this.)

Use mathematical induction to prove that $a_n = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$ for all integers $n \geq 1$.

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Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

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Solutions

Proof by Mathematical Induction

Base Case ($n=1$):

LHS: $a_1 = \sqrt{2}$ (given)

RHS: $2\cos\left(\frac{\pi}{2^{1+1}}\right) = 2\cos\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$

Since LHS = RHS, the statement holds for $n=1$.

Inductive Hypothesis:

Assume the statement is true for $n=k$, where $k$ is a positive integer:

$$a_k = 2\cos\left(\frac{\pi}{2^{k+1}}\right)$$

Inductive Step:

We must prove the statement for $n=k+1$:

$$a_{k+1} = 2\cos\left(\frac{\pi}{2^{k+2}}\right)$$

Using the recurrence relation $a_{k+1}^2 = 2 + a_k$ and the inductive hypothesis:

$$\begin{aligned} a_{k+1}^2 &= 2 + 2\cos\left(\frac{\pi}{2^{k+1}}\right) \\ &= 2\left[1 + \cos\left(\frac{\pi}{2^{k+1}}\right)\right] \end{aligned}$$

We now recognise that the expression inside the brackets is of the form $1+\cos(2\theta)$, so we can apply the double angle identity to rewrite it as a perfect square.

Apply the double angle identity: $1 + \cos(2\theta) = 2\cos^2(\theta)$.

Let $2\theta = \frac{\pi}{2^{k+1}}$, so $\theta = \frac{\pi}{2^{k+2}}$:

$$\begin{aligned} a_{k+1}^2 &= 2 \cdot 2\cos^2\left(\frac{\pi}{2^{k+2}}\right) \\ &= 4\cos^2\left(\frac{\pi}{2^{k+2}}\right) \end{aligned}$$

Taking square roots yields two possible signs:

$$a_{k+1} = \pm 2\cos\left(\frac{\pi}{2^{k+2}}\right)$$

We must determine the correct sign by examining the range of the angle and the definition of $a_n$. Since $n \geq 1$, we have $0 < \frac{\pi}{2^{k+2}} < \frac{\pi}{2}$, so $\cos\left(\frac{\pi}{2^{k+2}}\right) > 0$.

Also, $a_n$ is defined as nested square roots of positive numbers, so $a_{k+1} > 0$.

Therefore:

$$a_{k+1} = 2\cos\left(\frac{\pi}{2^{k+2}}\right)$$

This completes the inductive step. \checkmark

Conclusion:

By mathematical induction, $a_n = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$ for all integers $n \geq 1$.

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Takeaways

• Induction with Recurrence: Use given recurrence relation in inductive step to relate $a_{k+1}$ to $a_k$
• Double Angle Formula: Identity $1 + \cos(2\theta) = 2\cos^2(\theta)$ is key to converting sum to square
• Sign Consideration: Must justify taking positive square root using domain/range analysis
• Angle Halving: Pattern shows $a_n$ relates to $\cos$ of successively halved angles ($\pi/4, \pi/8, \pi/16, \ldots$)

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Polynomials, HSC Probability