关于三角函数恒等式(sin(2m+1)θ)/sinθ展开式的疑问与求助

Hey there! I totally get why this identity might not feel "obvious" at first glance—let’s unpack it step by step, and I’ll show you why textbooks often frame it that way, plus why Taylor series isn’t the right tool here.

Why the Identity Holds (No Taylor Series Needed!) #

The simplest way to derive this is using complex numbers and geometric series (a standard trick in trigonometry for summing cosine/sine series). Here’s how it works:

1. Start with Euler’s formula: $e^{ik\theta} = \cos k\theta + i\sin k\theta$. Remember that $\cos(-k\theta) = \cos k\theta$, so $2\cos2k\theta = e^{i2k\theta} + e^{-i2k\theta}$.
2. Rewrite the right-hand side (RHS) of your identity as a sum of complex exponentials:
   $$
   1 + 2\cos2\theta + 2\cos4\theta + \dots + 2\cos2m\theta = \sum_{k=-m}^m e^{i2k\theta}
   $$
   (The $k=0$ term gives us the $1$, and each pair $k=\pm t$ gives $2\cos2t\theta$.)
3. This is a geometric series with first term $e^{-i2m\theta}$, common ratio $e^{i2\theta}$, and $2m+1$ terms. Use the geometric series sum formula $\frac{a(r^n - 1)}{r - 1}$:
   $$
   \sum_{k=-m}^m e^{i2k\theta} = \frac{e^{{-i2m\theta}\left(e}{i2(2m+1)\theta} - 1\right)}{e^{i2\theta} - 1}
   $$
4. Simplify the numerator and denominator using Euler’s formula again:
   • Numerator: $e^{i2(m+1)\theta} - e^{-i2m\theta} = e^{{i\theta}\left(e}{i(2m+1)\theta} - e^{-i(2m+1)\theta}\right) = e^{i\theta} \cdot 2i\sin(2m+1)\theta$
   • Denominator: $e^{i2\theta} - 1 = e^{{i\theta}\left(e}{i\theta} - e^{-i\theta}\right) = e^{i\theta} \cdot 2i\sin\theta$
5. Cancel out the common terms ($e^{i\theta}$ and $2i$) from numerator and denominator, and you’re left with:
   $$
   \frac{\sin(2m+1)\theta}{\sin\theta}
   $$
   Which matches the left-hand side (LHS) of your identity.

Why Textbooks Call It "Obvious" #

This might seem like a few steps, but in contexts where you’ve already learned geometric series and Euler’s formula, this derivation is straightforward and follows standard patterns. Textbooks sometimes use "obvious" to mean "follows directly from tools we’ve already covered"—not that it’s instantly intuitive for everyone!

Alternative: Mathematical Induction #

If complex numbers aren’t your jam, you can also prove this with induction:

• Base case ($m=1$): $\frac{\sin3\theta}{\sin\theta} = 3 - 4\sin^2\theta = 1 + 2\cos2\theta$, which matches the RHS.
• Inductive step: Assume the identity holds for $m=k$, then show it holds for $m=k+1$ by adding $2\cos2(k+1)\theta$ to both sides and using trigonometric identities to simplify the LHS.

Either way, Taylor series is unnecessary here—this identity is about summing a finite trigonometric series, not expanding a function into an infinite power series.

内容的提问来源于stack exchange，提问作者mathpieuler