How can you derive $\sin(x) = \sin(x+2\pi)$ from the Taylor series for $\sin(x)$?

Question

\begin{eqnarray*} \sin(x) & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\\ \sin(x+2π) & = & x + 2\pi - \frac{(x+2π)^3}{3!} + \frac{(x+2π)^5}{5!} - \ldots \\ \end{eqnarray*}

Those two series must be equal, but how can you show that by only manipulating the series?

Answer 1

A fairly easy way to introduce $\pi$ in trigonometric functions defined by series is:

1. Define

$$e^z=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$

Then use Euler's formula to define $\sin$ and $\cos$:

$$\sin z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}$$ $$\cos z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}$$

2. Then prove that $\cos$ has a least positive root, which you call $\pi/2$. For this, you can show easily that $\cos 0>0$ and $\cos 2<0$ (the latter using majoration of the rest in the series, which is alternating).

3. Prove and use $e^{a+b}=e^ae^b$ (it's a Cauchy product) to derive similar identities for $\sin$ and $\cos$.

4. Use (2) and (3) to prove that $\sin$ and $\cos$ are $2\pi$-periodic.

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Here is the detailed derivation

First, define

$$e^z=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$

The series converges for all $z\in\Bbb C$ by the ratio test, thus it defines an entire function on the complex plane. It is $C^{\infty}$ on $\Bbb C$, and the restriction to real $z$ is real-valued and also $C^\infty$.

Putting $z=0$, you have $e^0=1$, and by differentiating the series, you get $\dfrac{\mathrm{d}e^z}{\mathrm{d}z}=e^z$.

Then you define

$$\cos z=\frac{e^{iz}+e^{-iz}}{2}=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}$$ $$\sin z=\frac{e^{iz}-e^{-iz}}{2i}=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}$$

And while we are at it,

$$\cosh z=\frac{e^{z}+e^{-z}}{2}=\sum_{n=0}^\infty\frac{z^{2n}}{(2n)!}$$ $$\sinh z=\frac{e^{z}-e^{-z}}{2}=\sum_{n=0}^\infty\frac{z^{2n+1}}{(2n+1)!}$$

Thus $\cos$ and $\cosh$ are even, while $\sin$ and $\sinh$ are odd.

Of course, the restriction of these functions to real $z$ are real-valued.

You have

$$e^{iz}=\cos z+i\sin z$$

And the derivatives $\sin'=\cos$ and $\cos'=-\sin$.

Notice also that the terms in the series of $e^x$ are increasing for increasing $x\geq0$, thus $x\rightarrow e^x$ is increasing for $x\geq0$ and you have $e^x\geq1+x$ for $x\geq0$, and $e^x\underset {x\rightarrow+\infty}\longrightarrow+\infty$.

Let $(a,b)\in\Bbb C^2$. Since the series of $e^z$ is absolutely convergent for all $z$, the following equality holds

$$e^ae^b=\sum_{i=0}^\infty \frac{a^i}{i!}\sum_{j=0}^\infty \frac{a^j}{j!}=\sum_{n=0}^{\infty} u_n$$

With

$$u_n=\sum_{k=0}^n\frac{a^kb^{n-k}}{k!(n-k)!}=\frac{1}{n!}\sum_{k=0}^n {n\choose k}a^kb^{n-k}=\frac{(a+b)^n}{n!}$$

Thus $e^ae^b=e^{a+b}$ for all complex $a,b$.

Thus you have $e^ze^{-z}=1$, and $e^z$ is never zero.

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Digression on the real exponential

Hence for real $x$, $e^x\neq0$, and since the function is $C^0$ (even $C^\infty$), its sign does not change, and $\forall x\in\Bbb R, e^x>0$.

Also, since $e^xe^{-x}=1$ and $e^x\underset {x\rightarrow+\infty}\longrightarrow+\infty$, you have $e^x\underset {x\rightarrow-\infty}\longrightarrow0$.

And since the derivative of $e^x$ is itself, the derivative is also always positive, and the exponential is increasing on $\Bbb R$.

You can conclude it's a bijection, and since $e^ae^b=e^{a+b}$ and $e^0=1$, this proves that the exponential is a group isomorphism between $(\Bbb R,+)$ and $(\Bbb R^\star_+,\cdot)$.

Call $\log$ the inverse isomorphism, defined on $\Bbb R^\star_+$, with $\log (ab)=\log(a)+\log(b)$ for all $a>0, b>0$. Also, using the formula of derivation of an inverse function, you have $\log'(x)=1/x$.

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Trigonometric identities

From $e^ae^b=e^{a+b}$ and using Euler's identity, you can derive the usual trigonometric (and hyperbolic trigonometry) identities. I'll show how on an example:

$$\cos a\cos b-\sin a\sin b=\frac{e^{ia}+e^{-ia}}{2}\frac{e^{ib}+e^{-ib}}{2}-\frac{e^{ia}-e^{-ia}}{2i}\frac{e^{ib}-e^{-ib}}{2i}$$ $$=\frac14\left[(e^{ia}+e^{-ia})(e^{ib}+e^{-ib})+(e^{ia}-e^{-ia})(e^{ib}-e^{-ib})\right]$$ $$=\frac{1}{4}\left[\left(e^{i(a+b)}+e^{i(a-b)}+e^{i(b-a)}+e^{-i(a+b)}\right)+\left(e^{i(a+b)}-e^{i(a-b)}-e^{i(b-a)}+e^{-i(a+b)}\right)\right]$$ $$=\frac{e^{i(a+b)}+e^{-i(a+b)}}{2}=\cos(a+b)$$

Likewise, you have $\sin(a+b)=\sin a\cos b+\sin b\cos a$, and a bunch of other formulas.

In particular, you have for all $z\in\Bbb C$:

$$\cos^2 z + \sin^2 z=\cos(z-z)=1$$ $$\cos 2z=\cos^2 z-\sin^2 z=2\cos^2 z-1$$

These are true for real $z$, and since the functions are then real-valued, you have $|\cos x|\leq 1$ and $|\sin x| \leq 1$ for all $x\in\Bbb R$.

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Definition of $\pi$

You have $\cos 0=1$ from the series definition, and

$$\cos 2=\sum_{n=0}^\infty (-1)^n\frac{2^{2n}}{(2n)!}$$

The series is alternating with decreasing term after $n=1$, thus

$$\cos 2<1-\frac{2^2}{2!}+\frac{2^4}{4!} = -\frac13 <0$$

Since $\cos$ is continuous, it has at least one root in $]0,2[$.

The series for $\sin x$ is also alternating for $0< x\leq 2$, and its general term is decreasing after $n=0$, thus for $x\in[0,2]$,

$$\sin x \geq x-\frac{x^3}{6}=x\left(1-\frac{x^2}{6}\right)$$

The RHS of the inequality has roots $0$ and $\pm\sqrt{6}$, and $\sqrt{6}>2$, thus for $x\in]0,2]$, $\sin x>0$.

Since $\cos'=-\sin$, you have that the function $\cos$ is decreasing on $]0,2[$.

Therefore, $\cos x=0$ has one and only one root in $[0,2]$. Let's call this root $\frac{\pi}2$.

We have then $\cos \frac{\pi}2=0$, thus $\cos^2 \frac{\pi}2+ \sin^2 \frac{\pi}2=1$ implies $\sin \frac{\pi}2=\pm1$, and since it's positive, $\sin \frac{\pi}2=1$.

Also, from $\cos 2x=2\cos^2x-1$, you get that $\cos \pi=-1$, and then $\sin\pi=0$.

Notice that you have also

$$e^{i\pi}=\cos \pi+i\sin\pi=-1$$

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Trigonometric functions are periodic

From the identities

$$\cos (a+b)=\cos a\cos b - \sin a \sin b$$ $$\sin (a+b)=\sin a\cos b + \cos a \sin b$$

You get

$$\cos (a+\pi)=\cos a\cos \pi - \sin a\sin \pi=-\cos a$$ $$\sin (a+\pi)=\sin a\sin \pi + \cos a\sin \pi=-\sin a$$

And finally

$$\cos (a+2\pi)=\cos a$$ $$\sin (a+2\pi)=\sin a$$

Thus $\cos$ and $\sin$ are $2\pi$-periodic. We have still to prove it's the smallest possible period, but before, let's have a look at variations of $\cos$ and $\sin$ on one period $[0,2\pi]$.

We already know that for $x\in[0,\pi/2]$, $\cos x\geq 0$ and $\sin x\geq 0$, where the former is decreasing from $1$ to $0$, and the latter is increasing from $0$ to $1$.

First, we complete an half-period. Using the previous identities:

$$\cos (\pi-x)=-\cos x$$ $$\sin (\pi-x)=\sin x$$

Thus for $x \in [\pi/2,\pi]$, $\cos$ is decreasing from $0$ to $-1$, and $\sin x$ is decreasing from $1$ to $0$.

Then we complete the full period with

$$\cos (a+\pi)=-\cos a$$ $$\sin (a+\pi)=-\sin a$$

This means that for $x\in[2\pi]$, the only roots of $\cos x$ are $\pi/2$ and $3\pi/2$, and the only roots of $\sin x$ are $0$, $\pi$ and $2\pi$.

Now, is $2\pi$ the smallest period? Suppose there is a $\lambda \in ]0,2\pi[$ such that for all $a$, $\cos (a+\lambda)=\cos a$, then

$$\cos(a+\lambda)=\cos a\cos\lambda-\sin a\sin \lambda=\cos a$$

And for $a=\pi/2$,

$$-\sin \lambda=0$$

Thus $\lambda=\pi$, but then $\cos a=\cos(a+\lambda)=-\cos a$, which is not true for example for $a=0$. Thus $2\pi$ is the minimal period.

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What next?

You could define $\tan x=\frac{\sin x}{\cos x}$ and derive identities, then define inverse trigonometric functions on some wise restriction (since a periodic function has no inverse), and also define $a^b=e^{b\log a}$. And you have a construction of all so-called elementary functions.

Answer 2

With series manipulation, we can get $$ \begin{align} \sin(x+y) &=\sum_{k=0}^\infty(-1)^k\frac{(x+y)^{2k+1}}{(2k+1)!}\tag{1a}\\ &=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}\frac{(-1)^k}{(2k+1)!}\binom{2k+1}{j}x^jy^{2k+1-j}\tag{1b}\\ &=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}(-1)^k\frac{x^j}{j!}\frac{y^{2k+1-j}}{(2k+1-j)!}\tag{1c}\\ &=\sum_{k=0}^\infty\sum_{j=0}^k(-1)^j\frac{x^{2j+1}}{(2j+1)!}(-1)^{k-j}\frac{y^{2k-2j}}{(2k-2j)!}\tag{1d}\\ &+\sum_{k=0}^\infty\sum_{j=0}^k(-1)^j\frac{x^{2j}}{(2j)!}(-1)^{k-j}\frac{y^{2k-2j+1}}{(2k-2j+1)!}\tag{1e}\\ &=\sum_{j=0}^\infty(-1)^j\frac{x^{2j+1}}{(2j+1)!}\sum_{k=0}^\infty(-1)^{k}\frac{y^{2k}}{(2k)!}\tag{1f}\\ &+\sum_{j=0}^\infty(-1)^j\frac{x^{2j}}{(2j)!}\sum_{k=0}^\infty(-1)^k\frac{y^{2k+1}}{(2k+1)!}\tag{1g}\\[6pt] &=\sin(x)\cos(y)+\cos(x)\sin(y)\tag{1h} \end{align} $$ Explanation:
$\text{(1a)}$: series definition
$\text{(1b)}$: binomial theorem
$\text{(1c)}$: break binomial coefficient into factorials
$\text{(1d)}$: add the odd powers of $x$ from $\text{(1c)}$
$\text{(1e)}$: to the even powers of $x$ from $\text{(1c)}$
$\text{(1f)}$: change the order of summation and substitute $k\mapsto k+j$
$\text{(1g)}$: change the order of summation and substitute $k\mapsto k+j$
$\text{(1g)}$: series definitions

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We can prove that $\cos(2\pi)=1$ and $\sin(2\pi)=0$ using a couple of derivatives and some geometry. First, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\sin(x) &=\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}\\ &=\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\\[6pt] &=\cos(x)\tag{2} \end{align} $$ and $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\cos(x) &=\frac{\mathrm{d}}{\mathrm{d}x}\left(1+\sum_{k=1}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\right)\\ &=\sum_{k=1}^\infty(-1)^k\frac{x^{2k-1}}{(2k-1)!}\\ &=\sum_{k=0}^\infty(-1)^{k+1}\frac{x^{2k+1}}{(2k+1)!}\\[6pt] &=-\sin(x)\tag{3} \end{align} $$ Using $(2)$ and $(3)$, we get $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^2(x)+\cos^2(x)\right) &=2\sin(x)\cos(x)-2\cos(x)\sin(x)\\ &=0\tag{4} \end{align} $$ Equation $(4)$ says that $\sin^2(x)+\cos^2(x)$ is a constant, and since $\sin(0)=0$ and $\cos(0)=1$, we have $$ \sin^2(x)+\cos^2(x)=1\tag{5} $$ Thus, $(\cos(x),\sin(x))$ is on the unit circle and moves with speed $$ \begin{align} \left|\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x),\sin(x))\right| &=\left|\vphantom{\frac12}(-\sin(x),\cos(x))\right|\\[6pt] &=1\tag{6} \end{align} $$ To go around the unit circle once at speed $1$ takes $2\pi$. Thus, $$ \begin{align} (\cos(2\pi),\sin(2\pi)) &=(\cos(0),\sin(0))\\[6pt] &=(1,0)\tag{7} \end{align} $$

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Combining $(1)$ and $(7)$ yields $$ \begin{align} \sin(x+2\pi) &=\sin(x)\cos(2\pi)+\cos(x)\sin(2\pi)\\[6pt] &=\sin(x)\tag{8} \end{align} $$

Answer 3

Note that $\sin$ is the unique solution of $y'' = -y$ subject to $y(0) = 0, y'(0) = 1$. Note that $x \mapsto \sin (x+2 \pi)$ also satisfies the same differential equation, so the question boils down to showing that $\sin (2 \pi) = 0, \cos(2 \pi) = 1$ (where $\cos = \sin'$).

Consider $\eta(x) = \sin^2 x + \cos^2 x$, we see $\eta(0) = 1$ and $\eta(x)'= 0$, hence $\sin^2 x + \cos^2 x =1$ for all $x$, and so we only need to establish that $\cos (2 \pi) = 1$.

It is straightforward to establish that there is a smallest $T>0$ such that $ \cos T = 0$. Rudin has a nice method in "Real & Complex Analysis". Alternatively, suppose $\cos x >0$ for all $x \ge 0$. We have $\cos 0 = 1$, and so we have some $x_0>0$ such that $\sin x_0 >0$. By presumption, $\sin $ is strictly increasing on $x \ge 0$, and since $\cos' x = - \sin x$, we have $\cos' x \le - \sin x_0$ for all $x \ge x_0$, which contradicts $\cos x >0$ for all $x \ge 0$. Since $\cos$ is continuous, there is a smallest positive number $T$ such that $\cos T = 0$, and we can see that $\sin T = 1$, since $\sin$ is increasing on $[0,T)$.

We define $\pi = 2 T$.

Let $\zeta(x) = \cos ( {\pi \over 2}-x)$. We note that $\zeta(0) = 0, \zeta'(0) = 1$ and $\zeta'' = - \zeta$, and so by uniqueness we have $\sin x= \cos ({\pi \over 2}-x) = \cos (x-{\pi \over 2})$ for all $x$ (since $\cos $ is even), and differentiating gives $\cos x = -\sin (x-{\pi \over 2})$ (since $\cos' = -\sin$). It is straightforward to establish that $\cos (2 \pi) = 1$ from this.

Addendum: To show that the solution of $y''+y = 0$ is unique we could use the Picard Lindelöf theorem. Alternatively, suppose $y_1,y_2$ are two solutions with the same initial conditions and let $\delta = y_1-y_2$. Let $V = \delta^2 + (\delta')^2$ and note that $V' = 0$. Since $V(0) = 0$, we have $V(x) = 0$ for all $x$, that is, $y_1 = y_2$.

Answer 4

Define $f(x)$ as the residual: $$\sin(x+2\pi) = x + 2\pi - \frac{(x+2\pi)^3}{3!} + \frac{(x+2\pi)^5}{5!}+\ldots$$ $$\equiv x - \frac{x^3}{3!} + \frac{x^5}{5!}+\ldots + f(x)=\sin(x)+f(x)$$ Now prove it is zero: $$f(x) = 2\pi -\frac{1}{3!}\left((2\pi)^3 + 3(2\pi)^2 x + 3(2\pi)x^2 \right)+ \frac{1}{5!}\left( (2\pi)^5 +5x (2\pi)^4+10x^2(2\pi)^3+10x^3(2\pi)^2+5(2\pi)x^4\right)+\ldots$$ Extract all powers of $x$ lower than $2$, and define yet another residual, $g(x)$: $$f(x)=\sin(2\pi) + x\sum_{k=1}^\infty \frac{(-1)^k(2\pi)^{2k}}{(2k)!}+g(x)$$ $$=\sin(2\pi) +x(1-\cos(2\pi))+g(x)$$ The first terms are zero, and now we must prove that the remaining residual $g(x)$ is also zero, ad infinitum.

A more general continuation to follow...

Answer 5

Using term-wise derivation, we define the auxiliary sum $$e^{ix}=\sin'(x)+i\sin(x)=\sum_{k=0}^\infty\frac{(ix)^k}{k!}.$$ By a straightforward manipulation of this sum using the Binomial theorem,

$$e^{i(x+t)}=\sum_{k=0}^\infty\frac{(i(x+t))^k}{k!}=\sum_{k=0}^\infty\sum_{l=0}^k\frac{k!i^kx^lt^{k-l}}{k!l!(k-l)!}=\sum_{l=0}^\infty\sum_{j=0}^\infty\frac{(ix)^l(it)^{j}}{l!j!}=e^{ix}e^{it}.$$

Then we evaluate $$\sin(1)=1-\frac1{3!}+\text{ positive pairs of terms }\cdots>0$$ $$\sin(4)=4-\frac{4^3}{3!}+\frac{4^5}{5!}-\frac{4^7}{7!}+\frac{4^9}{9!}-\text{ positive pairs of terms }\cdots<0,$$ to show that $\sin$ has a root in $(1,4)$, let $\pi$.

As $|e^{i\pi}|^2=e^{i\pi}\left(e^{i\pi}\right)^*=e^{i\pi}e^{-i\pi}=e^0=1$, this root is such that $$e^{i\pi}=\pm1,$$ and to avoid the possible negative, we use $$e^{i2\pi}=\left(e^{i\pi}\right)^2=1.$$Thus there is a real $\pi\in(1,4)$ such that $$\sin(x+2\pi)=\Im\left(e^{ix+i2\pi}\right)=\Im\left(e^{ix}\right)=\sin(x).$$