How to rewrite $\cos{2n\theta}$ as a summation of $\sin\theta$

Question

I want to rewrite $\cos{2 n \theta}$ as $$ \cos{2 n \theta}=\sum_{m=0}^{M} a_m \sin^m\theta $$ How to determine $M$ and coefficients $a_m$. Any comment is much appreciated. Many thanks in advance.

Answer 1

You can use that $$\begin{align} \cos(2n\theta)+i\sin(2n\theta)&=(\cos\theta+i\sin\theta)^{2n}\\ &=(\cos^2\theta-\sin^2\theta+2i\cos\theta\sin\theta)^n\\ &=(1-2\sin^2\theta+i\sin2\theta)^n \end{align}$$ and expand using the binomial theorem. Then (thanks to Michael Burr for the suggestion) all the $\sin(2\theta)$ of the real part appear with an even power, so you can substitute the identity $$\sin^22\theta=1-(1-2\sin^2\theta)^2.$$

Answer 2

This appears as an exercise in G. H. Hardy's A Course of Pure Mathematics, 10th edition, on page 397.

Let us start with an integral $$J_m=\int_0^x \sin^m t \sin a(x-t) \, dt\tag{1}$$ where $m$ is a non-negative integer and $$J_0=\int_0^x \sin a(x-t) \, dt=\frac{1-\cos ax} {a} \tag{2}$$ Now let $m\geq 2$ and we can write (via integration by parts) $$J_m=\frac{\sin^m x} {a} -\frac{m} {a} \int_0^x\sin^{m-1}t\cos t\cos a(x-t) \, dt$$ A further application of integration by parts gives us $$J_m=\frac{\sin^m x} {a} - \frac{m} {a^2} \int_0^x(-\sin^mt+(m-1)\sin^{m-2}\cos^2t)\sin a(x-t) \, dt$$ Rewriting $\cos^2t=1-\sin^2t$ in above integral we get $$J_m=\frac{\sin^mx} {a} + \frac{m} {a^2}J_m-\frac{m(m-1)}{a^2}J_{m-2}+\frac{m(m-1)}{a^2}J_m$$ ie $$m(m-1)J_{m-2}=a\sin^mx+(m^2-a^2)J_m\tag{3}$$ Putting $m=2,4,6, \dots$ we get $$2\cdot 1\cdot J_0=a\sin^2x+(2^2-a^2)J_2\\4\cdot 3 \cdot J_2=a\sin^4x+(4^2-a^2)J_4\\6\cdot 5\cdot J_4=a\sin^6x+(6^2-a^2)J_6\\ \dots$$ Noting the value of $J_0$ from $(2)$ we get $$\cos ax=1-\frac{a^2}{2!}\sin^2x+\frac{a^2(a^2-2^2)}{4! }\sin^4x-\frac{a^2(a^2-2^2)(a^2-4^2)}{6!}\sin^6x+\dots$$ The above series terminates when $a$ is an even integer (applicable to current question with $a=2n$) otherwise it is an infinite convergent series.

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Another approach is to deal with the function $f(x) =\cos(a\arcsin x) $ and find its Taylor series. To that end we first find the differential equation for $y=f(x)$.

We have $$y'=-\frac{a} {\sqrt {1-x^2}}\sin(a\arcsin x) $$ or $$\sqrt{1-x^2}y'+a\sin(a\arcsin x) =0$$ Differentiating again we get $$\sqrt{1-x^2}y''-\frac{xy'}{\sqrt {1-x^2}}+\frac{a^2}{\sqrt {1-x^2}}\cos(a\arcsin x) =0$$ or $$(1-x^2)y''-xy'+a^2y=0\tag{4}$$ Differentiating the above $n$ times using Leibniz rule (and using notation $y_n=f^{(n)} (x)$) we get $$(1-x^2)y_{n+2}-2nxy_{n+1}-n(n-1)y_n-xy_{n+1}-ny_n+a^2y_n=0$$ or $$(1-x^2)y_{n+2}-(2n+1)xy_{n+1}+(a^2-n^2)y_n=0\tag{5}$$ It should be observed that the function $y=\sin(a\arcsin x) $ also satisfies the same differential equations $(4),(5)$. Putting $x=0$ in $(5)$ we get $$y_{n+2}(0)=-(a^2-n^2)y_n(0)$$ Noting that $y(0)=1,y_1(0)=0$ we get via Taylor's theorem $$\cos(a\arcsin x) =f(x) =1-\frac{a^2}{2!}x^2+\frac{a^2(a^2-2^2)}{4!}x^4-\dots$$ Replacing $x$ by $\sin x$ we get the desired series for $\cos(ax) $ in powers of $\sin x$. The issues of convergence/sum of the above series can be handled by using integral form of remainder in Taylor's theorem and one can prove that the formula holds for all values of $a, x$.

Using similar approach for $f(x) =\sin(a\arcsin x) $ we can get the series $$\sin(ax) =a\sin x-\frac{a(a^2-1^2)} {3!}\sin^3x+\frac{a(a^2-1^2)(a^2-3^2)}{5!} \sin^5x-\dots $$ The above series can also be obtained using purely algebraic manipulation and this approach was mentioned by S. L. Loney in Plane Trigonometry and has been discussed in this answer.

Answer 3

Expanding De Moivre's theorem directly leads to $$ \begin{aligned} \cos(2n\theta)+i\sin(2n\theta)&=(\cos\theta+i\sin\theta)^{2n} \\ &=\sum\limits_{r=0}^{2n}{2n \choose r}\cos^r\theta (i\sin\theta)^{2n-r} \\ \end{aligned} $$ Taking the real part, we find that only the even powers of $\cos\theta$ and $\sin\theta$ survive, which yields $$ \cos(2n\theta)=\sum\limits_{r=0}^{n}(-1)^{n-r}{2n \choose 2r}\cos^{2r}\theta\sin^{2n-2r}\theta $$ Substitute $\sin\theta$ for $\cos\theta$ $$ \begin{aligned} \cos(2n\theta)&=\sum\limits_{r=0}^{n}(-1)^{n-r}{2n \choose 2r}(1-\sin^2\theta)^r\sin^{2n-2r}\theta \\ &=\sum\limits_{r=0}^{n}(-1)^{n-r}{2n \choose 2r}\sum\limits_{s=0}^r{r \choose s}(-1)^s\sin^{2s}\theta\sin^{2n-2r}\theta \end{aligned} $$ Collect the $\sin^{2m}\theta$ term with $s+n-r=m$, whose coefficient is $$ (-1)^m\sum\limits_{r=n-m}^{n}{2n \choose 2r}{r \choose r+m-n} $$ We can use the combinatorial identity $$ \sum\limits_{r=n-m}^{n}{2n \choose 2r}{r \choose r+m-n}=\frac{4n^2(4n^2-2)\cdots[4n^2-(2m-2)^2]}{(2m)!} $$

Putting all terms together, we finally obtain an explicit expansion of $\cos(2n\theta)$ in terms of $\sin^{2m}\theta$: $$ \cos(2n\theta)=\sum\limits_{m=0}^{n}(-1)^m\frac{4n^2(4n^2-2)\cdots[4n^2-(2m-2)^2]}{(2m)!}\sin^{2m}(\theta) $$