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I need to prove the following:

$$\cos\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{4}\right) \dots \cos\left(\dfrac{x}{2^{n}}\right) = \dfrac{\sin x}{2^{n}\sin\left(\frac{x}{2^{n}}\right)} $$

I'm stuck proving for $k = n + 1$ when you have $\cos\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{4}\right) \dots \cos\left(\dfrac{x}{2^{n+1}}\right)$. How to show this example?

amWhy
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Andrej
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    Hint: https://www.gettyimages.com/detail/illustration/view-royalty-free-illustration/520735903 – Oscar Lanzi May 16 '25 at 11:24
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    May be you forgot the principle of induction, you assume your identity holds for $n=k\ge 1$ then you must show it's true for $n=k+1$. Then only need to multiply your RHS with $\cos(\frac{x}{2^{k+1}})$ then use double angle formula. – OnTheWay May 16 '25 at 11:48
  • Checking the formula for $n=1$ can be helpful. – user May 16 '25 at 15:14
  • In that case is infinity, here is up to $n$. – Andrej May 16 '25 at 17:30

1 Answers1

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Proving something for all natural numbers by induction means proving it for $n = 1$ and then proving it for $n = k + 1$ assuming it is true for $1 \le n \le k$. To do the first step, show

$$\cos\left(\frac{x}{2}\right) = \frac{\sin(x)}{2 \sin (\frac{x}{2})}.$$

This can be done using the trig half-angle formulas (or alternatively, see that $2\sin(\frac{x}{2}) = \frac{\sin(x)}{\cos(\frac{x}{2})}$ by the double angle formula for $\sin$). For the second step, you must show that

\begin{align*} \cos(\dfrac{x}{2})\cos(\dfrac{x}{4}) \cdots \cos(\dfrac{x}{2^{n}}) = \dfrac{\sin x}{2^{n}\sin(\tfrac{x}{2^{n}})} \implies \\\cos(\dfrac{x}{2})\cos(\dfrac{x}{4}) \cdots \cos(\dfrac{x}{2^{n + 1}}) = \dfrac{\sin x}{2^{n + 1}\sin(\tfrac{x}{2^{n + 1}})} \end{align*}

For this, note that given the antedecent, the consequent resolves to

$$\frac{\sin(x)}{2^n\sin(\frac{x}{2^n})} \cdot \cos\left(\dfrac{x}{2^{n + 1}}\right) = \dfrac{\sin x}{2^{n + 1}\sin(\tfrac{x}{2^{n + 1}})}$$

which again can be proven using half angle formulas, or note that since

$$2\cos\left(\dfrac{x}{2^{n + 1}}\right)\sin\left(\dfrac{x}{2^{n + 1}}\right) = \sin\left(\dfrac{x}{2^{n}}\right)$$

by the double angle formula for sine, we see that

$$\cos\left(\dfrac{x}{2^{n + 1}}\right) = \frac{\sin\left(\dfrac{x}{2^{n}}\right)}{2 \sin\left(\dfrac{x}{2^{n + 1}}\right)},$$

and making this substitution in the relevant equation proves the necessary statement.

Mathemagician314
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