I am trying to figure out the trig fact that for all $n$ $$\sin(2n+1)x = \text{a linear combination of odd powers of}\; \sin(x).$$

For example: $\sin(3x)=A\sin^3(x)+B\sin(x)$ and $\sin(5x)=A\sin^5(x)+B\sin^3(x)+c\sin(x)$.

I am struggling just to start this problem because I cannot identify any trig substitutions I can make here.

  • $\begingroup$ please use LaTeX formatting! $\endgroup$ – The Count Jan 17 '17 at 19:10
  • 2
    $\begingroup$ Hint: $\sin((2n+1)x)$ is an odd function. $\endgroup$ – Akiva Weinberger Jan 17 '17 at 19:13
  • $\begingroup$ Tutorial on LaTeX: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Akiva Weinberger Jan 17 '17 at 19:15
  • $\begingroup$ More explicitly, $\frac{\sin\big((n+1)x\big)}{\sin(x)}$ is the $n$-th Chebyshev polynomial of the second kind $U_n\big(\cos(x)\big)$. Note that $U_{2n}(t)$ is an even function, whence consists of only terms with even powers of $t$. Note also that $\cos^2(x)=1-\sin^2(x)$. $\endgroup$ – Batominovski Jan 17 '17 at 19:50

Let $x\in \mathbb{R}$, $z\in\mathbb{C}$ such that $z=\cos x + i \sin x$

Then $$z^{2k+1} = \cos(2k+1)x + i \sin (2k+1)x.$$

On the other hand: $$z^{2k+1}=\cos^{2k+1}x + i a_1\cos^{2k}x\sin x - b_1\cos^{2k-1}x\sin^2 x - ia_2\cos^{2k-2}\sin^3x + ... $$ where $a_i, b_i$ are some real numbers.

Comparing the imaginary parts we obtain: $$\sin (2k+1)x = a_1\cos^{2k}x \sin x - a_2\cos^{2k-2}\sin^3x+...$$ Note, that powers of sines are odd and powers of cosines are even. We can write cosines as: $$\cos^{2k}x = (1-\sin^2x)^k = 1 + c_1 \sin^2x+...+c_k \sin^{2k}x$$ so we obtain: $$\sin (2k+1)x = a_1 (1-\sin^2x)^k\sin x- a_2 (1-\sin^2x)^{k-1}\sin^3x+...$$ Now note, that in the equation above every sine is in the odd power.

Hope it helped.


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