Further Trigonometric Identities and Inverse Functions

Put $a:=\arcsin y$ and $b:=\arccos y$. Then $\sin a=y$, $\cos b=y$, and $a,b$ lie in the interiors of their respective principal intervals.

The interval placement of step 1.1, the monotonicity and special values in [L3], and evenness of cosine give $\cos a>0$ and $\sin b>0$. The Pythagorean identity then gives $\cos a=\sqrt{1-y^2}$ and $\sin b=\sqrt{1-y^2}$.

Apply [L6] to sine on $[-\pi \mathrm{/}2,\pi \mathrm{/}2]$ at $a$: [L1] supplies injectivity and [L2] supplies continuity. Since its derivative there is $\cos a\mathrm{\ne }0$, the inverse is differentiable at $y$ with $(\arcsin y{)}^{\mathrm{\prime }}=1\mathrm{/}\cos a=1\mathrm{/}\sqrt{1-y^2}$.

Apply [L6] to cosine on $[0,\pi ]$ at $b$: [L1] supplies injectivity and [L2] supplies continuity. Its derivative is $-\sin b\mathrm{\ne }0$, so $(\arccos y{)}^{\mathrm{\prime }}=-1\mathrm{/}\sin b=-1\mathrm{/}\sqrt{1-y^2}$.

Steps 3.1 and 3.2 prove the two derivative formulas.

By [L4], cosine is positive on $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$, so tangent is defined and continuous there. By [L1], [L2], and [L6], its derivative is positive and the restriction is strictly increasing, hence injective.

Fix $y\in \mathbb{R}$ and put $c:=\frac{1}{\sqrt{1+y^2}},s:=\frac{y}{\sqrt{1+y^2}}\mathrm{.}$ The radicand is positive, $c>0$, and $c^2+s^2=1$. Thus [L3] supplies a unique $t\in [0,2\pi )$ with $(\cos t,\sin t)=(c,s)$.

Since $\cos t=c>0$, [L4] places $t$ in $[0,\pi \mathrm{/}2)\cup (3\pi \mathrm{/}2,2\pi )$. Set $u:=t$ in the first case and $u:=t-2\pi$ in the second. Then $u\in (-\pi \mathrm{/}2,\pi \mathrm{/}2)$ and periodicity of sine and cosine gives $\tan u=s\mathrm{/}c=y$. Hence the restriction is surjective.

Step 1.1 gives continuity and injectivity, while step 2.1 gives surjectivity.

For $y\in \mathbb{R}$, put $u:=\arctan y$. Then $\tan u=y$ and [L3] gives $(\tan {)}^{\mathrm{\prime }}(u)=1+y^2>0$. Applying [L2] to the principal branch proves $(\arctan y{)}^{\mathrm{\prime }}=1\mathrm{/}(1+y^2)$.

The function $t\mapsto 1\mathrm{/}(1+t^2)$ is continuous. By [L4], its oriented integral from $0$ to $x$ has derivative $1\mathrm{/}(1+x^2)$ and value $0$ at $x=0$. By [L3], $\tan 0=0$, so the inverse identity in [L1] gives $\arctan 0=0$; step 1.1 gives its derivative and [L8] makes it continuous. Their difference is therefore continuous on $\mathbb{R}$ with zero derivative, so [L8] makes it zero.

If $\mathrm{\mid }t\mathrm{\mid }<1$, [L5] with $r=-t^2$ gives $\frac{1}{1+t^2}={\sum }_{n=0}^{\mathrm{\infty }}(-1{)}^n{t}^{2n}\mathrm{.}$ Termwise integration between $0$ and $x$ (reversing endpoints when $x<0$) and step 2.1 give the asserted arctangent series for $\mathrm{\mid }x\mathrm{\mid }<1$.

Let $S:={\sum }_{n\ge 0}(-1{)}^n\mathrm{/}(2n+1)$, which exists by [L6]. Abel's theorem and step 3.1 yield $S={\lim }_{x\uparrow 1}{\sum }_{n\ge 0}\frac{(-1{)}^n{x}^{2n+1}}{2n+1}={\lim }_{x\uparrow 1}\arctan x=\arctan 1.$ By [L7] and the principal range, $\arctan 1=\pi \mathrm{/}4$.

Steps 1.1–4.1 establish all four displayed claims.

The geometric, sine, cosine, logarithmic, and inverse-tangent identities, with exactly the displayed domains and endpoint assertions, are [L2]–[L5]. The exponential identity needs one further move, because the statement writes $e^x$, the real power of the base $e$ in the sense of [L16], while [L3] defines $\exp$ and defines $e$ only as $\exp (1)$. By [L17] every value of $\exp$ is positive, so $e=\exp (1)>0$ and [L16] assigns $e^x$ the value $\exp (x\log e)$. Since $\log$ is the inverse of $\exp$ [L16], $\log e=\log (\exp 1)=1$, so $e^x=\exp (x\cdot 1)=\exp (x)(x\in \mathbb{R}),$ and the series [L3] gives for $\exp (x)$ is therefore the series of $e^x$.

Suppose $f(x)={\sum }_{n\ge 0}{a}_n{x}^n$ for $\mathrm{\mid }x\mathrm{\mid }<R$ with $R>0$; this is a power series about $0$ in the sense of [L18], and the hypothesis is exactly that it converges, with sum $f(x)$, at every $x$ with $\mathrm{\mid }x\mathrm{\mid }<R$. Its radius of convergence $\rho$ is therefore at least $R$: by [L24] a power series diverges at every point farther from its centre than its radius, so a point of convergence $x$ has $\mathrm{\mid }x\mathrm{\mid }\le \rho$, and letting $\mathrm{\mid }x\mathrm{\mid }$ run over $[0,R)$ gives $\rho \ge R$. Every point of $(-R,R)$ thus lies strictly inside the radius, so by [L9] the sum is differentiable there and its derivative is again a power series of radius $\rho$; induction on $k$, using $(n+1)!=(n+1)n!$ from [L7] at each step, gives $f^{(k)}(x)={\sum }_{n\ge 0}\frac{(n+k)!}{n!}{a}_{n+k}{x}^n(\mathrm{\mid }x\mathrm{\mid }<R)$ for every $k\ge 0$, each again a power series about $0$ of radius $\rho$. Evaluating at the centre, [L18] gives $f^{(k)}(0)=k!a_k$. Every $f^{(k)}$ is differentiable on $(-R,R)$, hence continuous there by [L15], so $f$ is smooth on $(-R,R)$ in the sense of [L19] and [L1] applies at $a=0$: the Maclaurin series of $f$ is ${\sum }_{k\ge 0}{f}^{(k)}(0)x^k\mathrm{/}k!={\sum }_{k\ge 0}{a}_k{x}^k$. A power series representing $f$ near $0$ is therefore its Maclaurin series.

Fix $\alpha \in \mathbb{R}$ and define $c_0=1$ and $c_{n+1}=c_n(\alpha -n)\mathrm{/}(n+1)$ for $n\ge 0$.

On $(-1,1)$, the function $x\mapsto (1+x{)}^{-\alpha }$ is differentiable and has derivative $-\alpha (1+x{)}^{-\alpha -1}$. The inner map $x\mapsto 1+x$ is a polynomial, so by [L14] it is differentiable with derivative $1$ at every point of $(-1,1)$, each of which is a limit point of $(-1,1)$; its value $1+x$ lies in $(0,\mathrm{\infty })$ and is a limit point of $(0,\mathrm{\infty })$, where by [L6] the outer map $u\mapsto u^{-\alpha }$ is differentiable with derivative $-\alpha u^{-\alpha -1}$. The chain rule [L13] therefore gives the composite derivative $-\alpha (1+x{)}^{-\alpha -1}\cdot 1$.

If $\alpha =m$ is a nonnegative integer, the recurrence of step 1.3 gives $c_n=\left(\genfrac{}{}{0px}{}{m}{n}\right)$ for $0\le n\le m$ and $c_n=0$ for $n>m$. Indeed $c_0=1=\left(\genfrac{}{}{0px}{}{m}{0}\right)$ by [L20]; and if $n<m$ and $c_n=\left(\genfrac{}{}{0px}{}{m}{n}\right)=m!\mathrm{/}(n!(m-n)!)$ as a real number by [L12], then, since $(n+1)!=(n+1)n!$ and $(m-n)!=(m-n)(m-n-1)!$ with $m-n\ge 1$, $c_{n+1}=\frac{m!}{n!(m-n)!}\cdot \frac{m-n}{n+1}=\frac{m!}{(n+1)!(m-(n+1))!}=\left(\genfrac{}{}{0px}{}{m}{n+1}\right);$ while $c_{m+1}=c_m(m-m)\mathrm{/}(m+1)=0$, after which the recurrence keeps every term $0$. Taking $u=x$ and $v=1$ in the finite binomial theorem of [L7] then gives $(1+x{)}^m=(x+1{)}^m={\sum }_{k=0}^m\left(\genfrac{}{}{0px}{}{m}{k}\right)x^k{1}^{m-k}={\sum }_{k=0}^m\left(\genfrac{}{}{0px}{}{m}{k}\right)x^k={\sum }_{n\ge 0}{c}_n{x}^n,$ using $1^j=1$ for every natural $j$, which the recursion $1^0=1$, $1^{j+1}=1^j\cdot 1$ of [L7] gives by induction, and using $c_n=0$ for $n>m$. The exponent in $(1+x{)}^m$ is there the natural-number one of [L7]. For $\mathrm{\mid }x\mathrm{\mid }<1$ that value is also the real power $(1+x{)}^m$ of [L16]: since $1+x>0$, the real powers $(1+x{)}^n$ with $n\in \mathbb{N}$ satisfy $(1+x{)}^0=\exp (0\cdot \log (1+x))=\exp 0=1$, the value $\exp 0=1$ being the value at the centre of the series [L3] gives for $\exp$, by [L18]; and, by the addition law of [L16] together with $(1+x{)}^1=\exp (\log (1+x))=1+x$, also $(1+x{)}^{n+1}=(1+x{)}^n(1+x{)}^1=(1+x{)}^n(1+x)$. That is the recursion determining the natural-number powers in [L7], so the two readings of $(1+x{)}^m$ agree.

Suppose $\alpha$ is not a nonnegative integer and $0<\mathrm{\mid }x\mathrm{\mid }<1$. Then, by step 1.3, $c_0=1$ and no factor $\alpha -n$ of the recurrence vanishes, so every $c_n$ is nonzero, and $x^n\mathrm{\ne }0$; the ratios $q_n:=\mid c_{n+1}{x}^{n+1}\mathrm{/}(c_n{x}^n)\mid =\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }\alpha -n\mathrm{\mid }\mathrm{/}(n+1)$ are therefore defined. For every $n>\mathrm{\mid }\alpha \mathrm{\mid }$ one has $\mathrm{\mid }\alpha -n\mathrm{\mid }=n-\alpha$ and hence $q_n=\mathrm{\mid }x\mathrm{\mid }(1-\frac{\alpha +1}{n+1}),$ so, since $1\mathrm{/}(n+1)\to 0$ by [L23] and convergence of a sequence is a condition on its tails, [L23] gives $q_n\to \mathrm{\mid }x\mathrm{\mid }$. By [L22] the limit superior is that same limit, so ${\mathrm{lim\hspace{0.17em}sup}}_n{q}_n=\mathrm{\mid }x\mathrm{\mid }<1$ and [L8] makes ${\sum }_{n\ge 0}{c}_n{x}^n$ converge absolutely.

Thus the six series in step 1.1 are precisely the asserted Maclaurin expansions; the logarithmic and inverse-tangent endpoint values are values of the same series, not claims of an open interval beyond its radius.

Therefore, for every real $\alpha$, the power series $B(x):={\sum }_{n\ge 0}{c}_n{x}^n$ built from step 1.3 converges absolutely at every $x$ with $\mathrm{\mid }x\mathrm{\mid }<1$: when $\alpha =m$ is a nonnegative integer every term past index $m$ vanishes by step 2.1 and the series is a finite sum; at $x=0$ it converges to $c_0$ by [L18]; and every remaining case is step 2.2. Its radius of convergence in the sense of [L18] is therefore at least $1$.

By [L9] and step 3.1, $B$ is differentiable on $(-1,1)$ with $B^{\mathrm{\prime }}(x)={\sum }_{n\ge 0}(n+1)c_{n+1}{x}^n$, again a power series of radius at least $1$. The recurrence of step 1.3 gives $(n+1)c_{n+1}=(\alpha -n)c_n$, so $B^{\mathrm{\prime }}(x)={\sum }_{n\ge 0}(\alpha -n)c_n{x}^n$ for $\mathrm{\mid }x\mathrm{\mid }<1$. Both ${\sum }_{n\ge 0}\alpha c_n{x}^n$ and ${\sum }_{n\ge 0}n{c}_n{x}^n=x{\sum }_{n\ge 0}(n+1)c_{n+1}{x}^n$ converge there, the first by step 3.1 and the second because it is $x{B}^{\mathrm{\prime }}(x)$, so [L21] splits the sum termwise into $\alpha B(x)-x{B}^{\mathrm{\prime }}(x)$. Hence $B^{\mathrm{\prime }}(x)=\alpha B(x)-x{B}^{\mathrm{\prime }}(x)$, that is $(1+x)B^{\mathrm{\prime }}(x)=\alpha B(x)$, for $\mathrm{\mid }x\mathrm{\mid }<1$.

For $G(x):=(1+x{)}^{-\alpha }B(x)$, both factors are differentiable on $(-1,1)$ by step 1.4 and step 4.1, so the product rule gives $G^{\mathrm{\prime }}(x)=-\alpha (1+x{)}^{-\alpha -1}B(x)+(1+x{)}^{-\alpha }{B}^{\mathrm{\prime }}(x)$ there. On $(-1,1)$ the base $1+x$ is positive, so $(1+x{)}^1=\exp (\log (1+x))=1+x$ and the addition law at the real exponents $-\alpha -1$ and $1$ gives $(1+x{)}^{-\alpha -1}(1+x)=(1+x{)}^{-\alpha }$, both by [L16]. Multiplying the displayed derivative by $1+x$ and using $(1+x)B^{\mathrm{\prime }}(x)=\alpha B(x)$ from step 4.1 therefore gives $(1+x)G^{\mathrm{\prime }}(x)=-\alpha (1+x{)}^{-\alpha }B(x)+\alpha (1+x{)}^{-\alpha }B(x)=0$; since $1+x\mathrm{\ne }0$, $G^{\mathrm{\prime }}(x)=0$ throughout $(-1,1)$.