General formula for $\sin\left(k\arcsin (x)\right)$

Question

I'm wondering if there's a simple way to rewrite this in terms of $k$ and $x$, especially as a polynomial. It seems to me to crop up every so often, especially for $k=2$, when I integrate with trig substitution. But $k=2$ is not so bad, because I can use the double angle formula; it's the prospect of higher values of $k$ that motivates this question.

I think the law of sines may help? Or maybe even De Moivre's theorem, to find the length of the hypotenuse as the length of the angle changes, if we think of the right triangle drawn from $\arcsin(x)$ with side1 = $x$, hypotenuse = $1$, and side2 = $\sqrt{1 - x^2}$ as a complex number, though I'm not sure how that might work.

Answer 1

We have: $$\arcsin(x)=\frac{\pi}{2}-\arccos x$$ and: $$ \sin(m\arccos(x))=\sqrt{1-x^2} U_{m-1}(x), \qquad \cos(m\arccos x)=T_m(x), $$ where $T_m$ and $U_{m-1}$ are Chebyshev polynomials of the first and second kind, respectively.

So we simply have:

$$\sin(m\arcsin x)=\left\{\begin{array}{rcl}T_m(x)&\text{ if }&m\equiv 1\pmod{4},\\ \sqrt{1-x^2} U_{m-1}(x)&\text{ if }&m\equiv 2\pmod{4},\\-T_m(x)&\text{ if }&m\equiv 3\pmod{4},\\ -\sqrt{1-x^2} U_{m-1}(x)&\text{ if }&m\equiv 0\pmod{4}. \end{array}\right.$$

Answer 2

I would try to write $A[k](x)=\sin (k.\arcsin(x) )$ and $B[k](x)=\cos (k.\arcsin (x) )$

and then $A[k+1](x) = \sin ( \arcsin(x) + k.\arcsin (x) ) = x B[k](x) + \sqrt{1-x^2} A[k](x)$

You could then either use the same recurrence form for $B[k+1](x)$ and have a double recurrence relation, or use $B[k](x)=\sqrt{1-A[k](x)^2}$ and have a simple one

Answer 3

Funny coincidence: just saw this formula of Euler's on page 240 of the current (August 2014) Fibonacci Quarterly:

$\sin(ma) =\sum_{n=0}^{\infty}\dfrac{m}{(2n+1)!}(-1)^n\sin^{2n+1}(a) \prod_{k=1}^{n}(m^2-(2k-1)^2) $.

Putting $a = \sin^{-1}(x)$, so $x = \sin(a)$, this becomes

$\sin(m\sin^{-1}x) =\sum_{n=0}^{\infty}\dfrac{m}{(2n+1)!}(-1)^nx^{2n+1} \prod_{k=1}^{n}(m^2-(2k-1)^2) $.

Note that there are only a finite number of terms if $m$ is an odd integer.

A quite reasonable proof of Euler's formula is given in the paper.

(added later)

Here is the proof:

Let $f(a) = \sin(m \sin^{-1}(\sin(a)))$ or $f(x) = \sin(m \sin^{-1}(\sin(x)))$ .

Differentiating w.r.t. $x$, we get $f'(x) =m(1-x^2)^{-1/2} \cos(m \sin^{-1}(x)) $ and $f''(x) =m x (1-x^2)^{-3/2}\cos(m\sin^{-1}(x))-m^2(1-x^2)^{-1}\sin(m\sin^)-1)(x)) $ $ =x(1-x^2)^{-1}f'(x) - m^2(1-x^2)^{-1}f(x) $.

In other words, $f''(x)-P(x)f'(x)+Q(x)f(x) = 0$, where $P(x) = x/(1-x^2)$ and $Q(x) = m^2/(1-x^2)$. Since the denominators of $P$ and $Q$ are non-zero at $0$, these are analytic there. Thus, $x=0$ is an ordinary point of the differential equation $$(1-x^2)f''(x) - xf'(x) + m^2 f(x) = 0 $$ so there is a power series solution of the form $f(x) =\sum_{n=0}^{\infty} u_n x^n $.

Substituting this, we get $$u_{n+2} =-\dfrac{m^2-n^2}{(n+1)(n+2)}u_n , n \ge 0 . $$

Since $u_0 =f(0) =\sin(m \sin^{-1}(0)) = 0 $, $u_{2n} = 0$ for all $n$.

Since $u_1 =f'(0) =m(1-0^2)^{-1/2} = m $, $u_{2n+1} $ is given by the formula for all $n$.

Answer 4

Consider the differential equation $$ (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+k^2y=0 $$ for $k\in\mathbb{Z^+}$. Letting $y=\sum_{i=0}^{\infty}a_ix^i$ gives $$ \frac{dy}{dx}=\sum_{i=1}^\infty ia_ix^{i-1} $$ and $$ \frac{d^2y}{dx^2}=\sum_{i=2}^\infty i(i-1)a_ix^{i-2}. $$ Now substituting we get $$ \begin{align} (1-x^2)\sum_{i=2}^\infty i(i-1)a_ix^{i-2}-x\sum_{i=1}^\infty ia_ix^{i-1}+k^2\sum_{i=0}^{\infty}a_ix^i&=0 \\\iff \sum_{i=2}^\infty i(i-1)a_ix^{i-2}-\sum_{i=2}^\infty i(i-1)a_ix^{i}-\sum_{i=1}^\infty ia_ix^{i}+k^2\sum_{i=0}^{\infty}a_ix^i&=0 \\\iff \sum_{i=0}^\infty (i+2)(i+1)a_{i+2}x^{i}-\sum_{i=2}^\infty i^2a_ix^{i}+\sum_{i=2}^\infty ia_ix^{i}-\sum_{i=1}^\infty ia_ix^{i}+k^2\sum_{i=0}^{\infty}a_ix^i&=0 \\\iff (2a_2+k^2a_0)+(6a_3+k^2a_1-a_1)x+\sum_{i=2}^{\infty} ((i+2)(i+1)a_{i+2}+(k^2-i^2)a_i)x^i&=0 \\\iff\sum_{i=0}^{\infty} ((i+2)(i+1)a_{i+2}+(k^2-i^2)a_i)x^i&=0. \end{align} $$ As this is an identity, we need each coefficient to be 0 giving $$ a_{i+2}=-\frac{(k+i)(k-i)}{(i+2)(i+1)}a_i $$ $\forall i \geqslant 0$. At this point, we can freely choose $a_0,a_1\in\mathbb{R}$ as seed values, where each choice would yield a unique solution, subsequently leading to expressions for every even and odd coefficient, respectively. The specific solution we require is such that when $x=0, y=\sin(k\arcsin(0))=0$ and $\frac{dy}{dx}=\frac{kcos(k\arcsin(0))}{\sqrt{1-0^2}}=k$, so if this is indeed the case, we have $$ \begin{align} a_0+\sum_{i=1}^{\infty}a_i(0)^{i}&=a_0=0 \\a_1+\sum_{i=2}^{\infty}ia_i(0)^{i-1}&=a_1=k \end{align} $$ from the power series for $y$ evaluated at $x=0$. Now we have $$ \begin{align} a_{2}=-\frac{(k+0)(k-0)}{(0+2)(0+1)}a_0=0 \\\implies a_{4}=-\frac{(k+2)(k-2)}{(2+2)(2+1)}a_2=0 \\\vdots \\\implies a_{2n}=0 \end{align} $$ $\forall n\geqslant0$ and setting $i=2n-1$, $$ \begin{align} a_{2n+1}&=-\frac{(k+(2n-1))(k-(2n-1))}{((2n-1)+2)((2n-1)+1)}a_{2(n-1)+1} \\&=\cdots \\&=a_1\prod_{l=1}^{n}\left(-\frac{(k+(2l-1))(k-(2l-1))}{((2l-1)+2)((2l-1)+1)}\right) \\&=k(-1)^n\left(\prod_{l=1}^{n}(k+(2l-1))(k-(2l-1))\right)\left(\prod_{l=1}^{n}(2l+1)(2l)\right)^{-1} \\&=\frac{k(-1)^n}{(2n+1)!}\prod_{l=1}^{n}(k+(2l-1))(k-(2l-1)) \\&=\begin{cases} \displaystyle\frac{k(-1)^n}{(2n+1)!}\prod_{l=1}^{n}(k+(2l-1))(k-(2l-1)) , & k>2n-1\ \lor \ k\equiv0\pmod{2} \\ 0, & k \leqslant 2n-1 \ \land \ k\equiv1\pmod{2} \end{cases} \end{align} $$ $\forall n\geqslant 0$. Note that all the odd coefficients are non-zero when $k$ is even, as clearly $k-(2l-1)$ could never be $0$, i.e. we obtain an infinite power series. However, whenever $k=2j-1$ for some $j$, we have $a_{2n+1}=0 \ \forall n \geqslant j=\frac{k+1}{2}$, as $2j-1-(2l-1)=2(j-l)$ would be $0$ when $l=j$ (which will be achieved for $n\geqslant j$ as $1\leqslant l\leqslant n$). In this case we obtain a polynomial. So $$ \sin(k\arcsin x)= \begin{cases} \displaystyle \sum_{n=0}^\infty \frac{k(-1)^n}{(2n+1)!}\left(\prod_{l=1}^{n}(k+(2l-1))(k-(2l-1))\right)x^{2n+1}, & k\equiv0\pmod{2} \\\displaystyle\sum_{n=0}^k \frac{k(-1)^n}{(2n+1)!}\left(\prod_{l=1}^{n}(k+(2l-1))(k-(2l-1))\right)x^{2n+1}, & k\equiv1\pmod{2} \end{cases} $$