What is $\sin(nx)$ iteration in terms of $\sin A$ and $\cos A$?
I want to use sum angle formulas, $\sin(A+B)=\sin A\cos B+\cos A\sin B$ to get for any angles, $\sin(nA)$ in terms of powers of $\sin A$ and $\cos A$.
$\endgroup$ 4I know there are other ways, but I want to use trigonometry and iteration on that. The goal is to carry out iteration in terms of $\sin(nA)$?
4 Answers
$\begingroup$Let's find out the first ones!
$$\sin(2x)=\sin(x+x)=2\sin(x)\cos(x)$$
I'm going to get the cosine of that too while we're at it.
$$\cos(2x)=\cos(x+x)=\cos(x)^2-\sin(x)^2$$
$$\sin(3x)=\sin(x+2x)=\sin(x)\cos(2x)+\sin(2x)\cos(x)=\sin(x)\cos(x)^2-\sin(x)^3+2\sin(x)\cos(x)^2=3\sin(x)\cos(x)^2-\sin(x)^3$$
$$\cos(3x)=\cos(x+2x)=\cos(x)\cos(2x)-\sin(x)\sin(2x)=\cos(x)^3-\sin(x)^2\cos(x)-2\sin(x)^2\cos(x)=\cos(x)^3-3\sin(x)^2\cos(x)$$
Now repeat the procedure until you see a pattern:
$$\sin(4x)=4\sin(x)\cos(x)^3-4\sin(x)^3\cos(x)$$ $$\cos(4x)=\cos(x)^4-6\sin(x)^2\cos(x)^2+\sin(x)^4$$ $$\sin(5x)=5\sin(x)\cos(x)^4-10\sin(x)^3\cos(x)^2+\sin(x)^5$$ $$\cos(5x)=\cos(x)^5-10\sin(x)^2\cos(x)^3+5\sin(x)^4\cos(x)$$ $$\sin(6x)=\color{red}{6}\sin(x)\cos(x)^5-\color{red}{20}\sin(x)^3\cos(x)^3+\color{red}{6}\sin(x)^5\cos(x)$$ $$\cos(6x)=\color{blue}{1}\cos(x)^6-\color{blue}{15}\sin(x)^2\cos(x)^4+\color{blue}{15}\sin(x)^4\cos(x)^2-\color{blue}{1}\sin(x)^6$$
The coefficients become particularly interesting. Let's look at Pascal's triangle. $$1$$ $$1\quad 1$$ $$1\quad 2\quad 1$$ $$1\quad 3\quad 3 \quad 1$$ $$1\quad 4\quad 6\quad 4\quad 1$$ $$1\quad 5\quad 10\quad 10\quad 5\quad 1$$ $$\color{blue}{1}\quad \color{red}{6}\quad \color{blue}{15}\quad \color{red}{20}\quad \color{blue}{15}\quad \color{red}{6}\quad \color{blue}{1}$$
So if we ignored the minus signs $\cos(6x)+\sin(6x)$ would be $(\cos(x)+\sin(x))^6$
Also note that the minus signs seem to alternate (I put the higher powers of $\cos(x)$ first so this happened). If you have done some problem solving with complex numbers this is a dead giveaway of powers of a root of unity.
If you look at the $\cos(nx)$ you'll see the minus signs seem to be accompanied by $\sin(x)^2$ but not by $\sin(x)^4$. I'm going to reproduce them so the colors don't get too dense:
$$\cos(2x)=\cos(x)^2\color{red}{-}\sin(x)^\color{red}{2}$$
$$\cos(4x)=\cos(x)^4\color{red}{-}6\sin(x)^\color{red}{2}\cos(x)^2+\sin(x)^4$$
$$\cos(6x)=1\cos(x)^6\color{red}{-}15\sin(x)^\color{red}{2}\cos(x)^4+15\sin(x)^4\cos(x)^2\color{red}{-}1\sin(x)^\color{red}{6}$$
So instead we might consider $\omega\sin(x)$ instead of $\sin(x)$ where $\omega^2=-1$ and $\omega^4=1$ and the imaginary unit does just this, how lucky we are. $$i^1=i$$ $$i^2=-1$$ $$i^3=-i$$ $$i^4=1$$
so
$$\cos(6x)=\cos(x)^6+15(i\sin(x))^2\cos(x)^4+15(i\sin(x))^4\cos(x)^2+(i\sin(x))^6$$
We weren't so sure before but now this definitely looks like a binomial expansion. Since we were considering $(\cos(x)+\sin(x))^6$ before, let's try $(\cos(x)+i\sin(x))^6$ now.
$$(\cos(x)+i\sin(x))^6=\cos(x)^6+6\cos(x)^5(i\sin(x))+15\cos(x)^4(i\sin(x))^2+20\cos(x)^3(i\sin(x))^3+15\cos(x)^2(i\sin(x))^4+6\cos(x)(i\sin(x))^5+(i\sin(x))^6$$
Since the even powers of $i$ are real and the odd ones are imaginary let's group them together.
$$(\cos(x)+i\sin(x))^6=\cos(x)^6+15\cos(x)^4\sin(x)^2+15\sin(x)^4\cos(x)^2-\sin(x)^6+i(6\sin(x)\cos(x)^5-20\sin(x)^3\cos(x)^3+6\sin(x)^5\cos(x))=\cos(6x)+i\sin(6x)$$
Verifying this for all previous powers is left as an exercise for the reader. Now we might conjecture that $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for all positive integers n. We can rigorously prove this! Since we're trying to prove something for all positive integers, induction is worth a try.
Our base case is $(\cos(x)+i\sin(x))^0=1$ since $\cos(0x)=1$ and $\sin(0x)=0$ it is always true that $(\cos(x)+i\sin(x))^0=\cos(0x)+i\sin(0x)$.
Now for the inductive step, if
$$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$$
$$(\cos(x)+i\sin(x))^{n+1}=(\cos(nx)+i\sin(nx))(\cos(x)+i\sin(x))$$
$$(\cos(x)+i\sin(x))^{n+1}=\cos(nx)\cos(x)-\sin(nx)\sin(x)+i(\sin(nx)\cos(x)+\cos(nx)\sin(x))$$
$$(\cos(x)+i\sin(x))^{n+1}=\cos(nx+x)+i\sin(nx+x)$$
$$(\cos(x)+i\sin(x))^{n+1}=\cos((n+1)x)+i\sin((n+1)x)$$
So the identity being true for $n$ implies it being true for $n+1$, so we conclude by induction that $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for all positive integers $n$.
If you want $\sin(nx)$ and $\cos(nx)$ explicitly you can open that binomial and separate the real and imaginary parts:
$$\cos(nx)+i\sin(nx)=\sum_{h=0}^n{n\choose h}\cos(x)^h(i\sin(x))^{n-h}$$
$$\cos(nx)=\sum_{h=0, 2\mid h}^n{n\choose h}\cos(x)^h(i\sin(x))^{n-h}$$
$$i\sin(nx)=\sum_{h=0, 2\not\mid h}^n{n\choose h}\cos(x)^h(i\sin(x))^{n-h}$$
This is clearly not as elegant.
You could eliminate the imaginary unity from the equation but that would be kind of ugly. For me this shows how the complex numbers are beautiful.
$\endgroup$ $\begingroup$Is this OK? Trigonometry Sine Multiple Angle Formulae from Ken Ward's Mathematics Pages.
$\endgroup$ $\begingroup$$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ You can get a recursive relation for $\ds{\sin\pars{nx}}$ because: $$ \sin\pars{\bracks{m + 1}x} + \sin\pars{\bracks{m - 1}x} = 2\sin\pars{mx}\cos\pars{x} $$ Set $\ds{m = n - 1}$ in the above expression to find: \begin{align} \bbox[#ffd,10px,border:1px dotted navy]{\sin\pars{nx}} & = -\sin\pars{\bracks{n - 2}x} + 2\sin\pars{\bracks{n - 1}x}\cos\pars{x} \\[5mm] & =\ \bbox[#ffd,10px,border:1px dotted navy]{% \sum_{k = 1}^{2}a_{k}\pars{x}\sin\pars{\bracks{n - k}x}} \qquad\mbox{where}\qquad \left\{\begin{array}{rcl} \ds{a_{1}\pars{x}} & \ds{=} & \ds{2\cos\pars{x}} \\[1mm] \ds{a_{2}\pars{x}} & \ds{=} & \ds{-1} \end{array}\right. \end{align}
For instance, \begin{align} \bbox[#ffd,10px,border:1px dotted navy]{\sin\pars{4x}} &\ = -\sin\pars{2x} + 2\sin\pars{3x}\cos\pars{x} \\[5mm] & = -\sin\pars{2x} + 2\bracks{-\sin\pars{x} + 2\sin\pars{2x}\cos\pars{x}}\cos\pars{x} \\[5mm] & = \bracks{4\cos^{2}\pars{x} - 1}\bracks{2\sin\pars{x}\cos\pars{x}} - 2\sin\pars{x}\cos\pars{x} \\[5mm] & =\ \bbox[#ffd,10px,border:1px dotted navy]{% 8\sin\pars{x}\cos^{3}\pars{x} - 4\sin\pars{x}\cos\pars{x}} \end{align} $\endgroup$ $\begingroup$
The angle addition formulas for sine and cosine are nicely captured by this matrix formula $$ \left( \begin{array}{c} s_{n} \\ c_{n}\end{array} \right) = \left( \begin{array}{r} c & s \\ -s & c\end{array} \right) \left( \begin{array}{c} s_{n-1} \\ c_{n-1}\end{array} \right) $$ where $s_k = \sin(kx)$, $c_k = \cos(kx)$, $s= s_1$ and $c=c_1$.
Playing this recursion backwards, we arrive at the formula $$ \left( \begin{array}{c} s_{n} \\ c_{n}\end{array} \right) = \left( \begin{array}{r} c & s \\ -s & c\end{array} \right)^n \left( \begin{array}{c} s_{0} \\ c_{0}\end{array} \right) $$
Note that $s_0=0$ and $c_0=1$.
It's worth pointing out that that $$ \left( \begin{array}{r} c & s \\ -s & c\end{array} \right) = c \left( \begin{array}{r} 1 & 0 \\ 0 & 1\end{array} \right) + s \left( \begin{array}{r} 0 & 1 \\ -1 & 0\end{array} \right) $$ and that $$ \left( \begin{array}{r} 0 & 1 \\ -1 & 0 \end{array}\right)^2 = \left( \begin{array}{r} -1 & 0 \\ 0 & -1\end{array} \right) $$
$\endgroup$
