Sine, Cosine, and the Definition of Pi

For the sine absolute terms, the successive ratio is $\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}((2n+2)(2n+3))$, which tends to $0$.

For the cosine absolute terms, the successive ratio is $\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}((2n+1)(2n+2))$, which tends to $0$.

The ratio test proves absolute convergence of both series for this arbitrary $x$.

Termwise differentiation of the sine series gives ${\sum }_{n\ge 0}(-1{)}^n{x}^{2n}\mathrm{/}(2n)!=\cos x$.

Termwise differentiation of the cosine series gives $-{\sum }_{n\ge 0}(-1{)}^n{x}^{2n+1}\mathrm{/}(2n+1)!=-\sin x$.

Evaluating the defining series at $0$ gives $\sin 0=0$ and $\cos 0=1$.

By [L1], $y=a\cos +b\sin$ satisfies $y^{\mathrm{\prime }\mathrm{\prime }}=-y$, $y(0)=a$, and $y^{\mathrm{\prime }}(0)=b$.

Put $h=z-y$. Then $h^{\mathrm{\prime }\mathrm{\prime }}=-h$, $h(0)=h^{\mathrm{\prime }}(0)=0$, and $E:=h^2+(h^{\mathrm{\prime }}{)}^2$ has derivative $E^{\mathrm{\prime }}=2h(h+h^{\mathrm{\prime }\mathrm{\prime }})=0$.

Thus $E$ is constantly $E(0)=0$; as both squares are nonnegative, $h=0$ everywhere and $z=y$.

The function $x\mapsto \sin (x+y)$ solves $u^{\mathrm{\prime }\mathrm{\prime }}=-u$ with initial values $u(0)=\sin y$, $u^{\mathrm{\prime }}(0)=\cos y$.

The function $x\mapsto \sin x\cos y+\cos x\sin y$ has the same equation and initial values.

Uniqueness in [L1] gives the sine addition formula.

Repeating the argument for $x\mapsto \cos (x+y)$ and $\cos x\cos y-\sin x\sin y$ gives the cosine addition formula.

The derivative of $F(x)={\sin }^{2}x+{\cos }^{2}x$ is $2\sin x\cos x-2\cos x\sin x=0$.

Step 1.1 makes $F$ differentiable everywhere, hence continuous by [L4]. The zero-derivative theorem therefore makes $F$ constant, and $F(0)=1$, proving ${\sin }^{2}x+{\cos }^{2}x=1$; each square is then at most $1$.

Applying the addition formulas at $x+(-x)=0$ gives $0=\sin x\cos (-x)+\cos x\sin (-x),1=\cos x\cos (-x)-\sin x\sin (-x)\mathrm{.}$ The coefficient matrix squares to the identity by step 2.1, so these equations give $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$.

The absolute sine terms after the first have successive ratio at most $4\mathrm{/}6<1$, so [L2] gives $\sin x\ge x-x^3\mathrm{/}6\ge x\mathrm{/}3>0$.

In the cosine series at $2$, the first three terms sum to $1-2+2\mathrm{/}3=-1\mathrm{/}3$, and the remaining alternating tail begins negative with decreasing absolute terms; hence $\cos 2\le -1\mathrm{/}3$.

On $(0,2)$ one has ${\cos }^{\mathrm{\prime }}=-\sin <0$ by step 1.1, so the mean value theorem makes $\cos$ strictly decreasing on $[0,2]$.

Continuity, $\cos 0>0$, and $\cos 2<0$ give some $\gamma \in (0,2)$ with $\cos \gamma =0$.

Strict decrease on $[0,2]$ makes this zero unique and gives $\cos x>0$ for $0\le x<\gamma$.

No positive number below $\gamma$ is a zero, so $\gamma$ is the smallest positive zero.

From ${\cos }^2\gamma +{\sin }^2\gamma =1$, $\cos \gamma =0$, and $\sin \gamma >0$, one gets $\sin \gamma =1$.

Substituting $\gamma$ into the addition formulas gives $\sin (x+\gamma )=\cos x$ and $\cos (x+\gamma )=-\sin x$.

Applying step 2.1 twice gives the shifts by $2\gamma =\pi$ and the listed special values.

The first assertion is in [L1].

If $0<x\le \gamma$, positivity follows from [L2]; if $\gamma <x<2\gamma$, write $x=\gamma +t$ with $0<t<\gamma$, and [L1] and [L2] give $\sin x=\cos t>0$.

These cases cover $0<x<\pi =2\gamma$, proving that $\pi$ is the first positive sine zero.

Natural induction applied to the $\pi$ shift gives $\sin (x+n\pi )=(-1{)}^n\sin x$ and $\cos (x+n\pi )=(-1{)}^n\cos x$ for every natural $n$. Applying these identities at $x-n\pi$ gives the matching backward shifts; since every integer is $n$ or $-n$ and $(-1{)}^{-n}=(-1{)}^n$, the displayed identities hold for every integer $m$.

Choose an integer $m$ with $m\pi \le x<(m+1)\pi$ and put $r=x-m\pi \in [0,\pi )$. By [L1] and step 1.1, $\sin x=0$ exactly when $r=0$, hence exactly when $x=m\pi$.

The quarter-turn shift converts the sine zero set into $\cos x=0$ exactly when $x=(m+1\mathrm{/}2)\pi$.

Step 1.1 with $m=2$ gives period $2\pi$. A positive common period $T$ has $\sin T=0$, hence $T=m\pi$ by step 2.1; $m=1$ fails for cosine because $\cos \pi =-1$, so $m\ge 2$.

Therefore $2\pi$ is the least positive common period.

On the open intervals where $\cos$ is positive respectively negative, ${\sin }^{\mathrm{\prime }}=\cos$ makes sine strictly increasing respectively decreasing.

On the open intervals where $\sin$ is positive respectively negative, ${\cos }^{\mathrm{\prime }}=-\sin$ makes cosine strictly decreasing respectively increasing.

The period moves these conclusions to every integer $m$, and the endpoint values in [L2] show both ranges are exactly $[-1,1]$.

Quotient and reciprocal differentiation applied to [L1] gives the four displayed derivatives after using ${\sin }^2+{\cos }^2=1$.

Shifting by $\pi$ negates both sine and cosine, so their quotients have period $\pi$, while their reciprocals change sign and therefore have period $2\pi$.

The zero-set and quarter-turn values rule out a smaller positive period in each case.

Substituting [L1] into the derivative definition [L2] gives ${\lim }_{x\to 0}\sin x\mathrm{/}x=1$.

This is the asserted limit.