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Section 8.4 Trigonometric Functions

We now construct sine and cosine from their power series. The basic algebraic identities will emerge from differentiation and from a simple uniqueness argument for the differential equation \(y''+y=0\text{.}\) This follows the same general strategy used in classical analysis texts such as Rosenlicht.

Subsection 8.4.1

For \(x \in \R\text{,}\) define
\begin{equation*} \sin x := \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} \qquad\text{and}\qquad \cos x := \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}. \end{equation*}

Proof.
If \(|x| \le R\text{,}\) then
\begin{equation*} \left|\frac{(-1)^n x^{2n+1}}{(2n+1)!}\right| \le \frac{R^{2n+1}}{(2n+1)!}, \qquad \left|\frac{(-1)^n x^{2n}}{(2n)!}\right| \le \frac{R^{2n}}{(2n)!}. \end{equation*}
Both numerical majorants converge by the ratio test, so the Weierstrass M-test (Theorem 8.1.6) gives uniform convergence on \([-R,R]\text{.}\)

Proof.
Each defining series has radius of convergence \(\infty\text{,}\) so Theorem 6.2.2 allows us to differentiate term by term:
\begin{equation*} (\sin x)' = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = \cos x, \end{equation*}
and
\begin{equation*} (\cos x)' = \sum_{n=1}^\infty \frac{(-1)^n (2n)x^{2n-1}}{(2n)!} = -\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = -\sin x. \end{equation*}
Analyticity follows because both are power-series functions with infinite radius of convergence.
The defining series immediately show that
\begin{equation*} \sin(-x)=-\sin x \qquad\text{and}\qquad \cos(-x)=\cos x, \end{equation*}
so \(\sin\) is odd and \(\cos\) is even.

Proof.
Let \(h(x)=\sin^2 x+\cos^2 x\text{.}\) Using the product rule and Proposition 8.4.2, we get
\begin{equation*} h'(x)=2\sin x \cos x + 2\cos x(-\sin x)=0. \end{equation*}
By Corollary 5.4.5, \(h\) is constant on \(\R\text{.}\) Evaluating at \(x=0\) gives
\begin{equation*} h(0)=\sin^2 0+\cos^2 0=0^2+1^2=1. \end{equation*}
Hence \(h(x)=1\) for all \(x\text{.}\)

Proof.
Consider the function
\begin{equation*} E(x)=h(x)^2+h'(x)^2. \end{equation*}
Differentiating gives
\begin{equation*} E'(x)=2h(x)h'(x)+2h'(x)h''(x) = 2h'(x)\bigl(h(x)+h''(x)\bigr) =0. \end{equation*}
Hence \(E\) is constant by Corollary 5.4.5. Since \(E(0)=h(0)^2+h'(0)^2=0\text{,}\) we have \(E(x)=0\) for every \(x\text{.}\) Therefore both \(h(x)\) and \(h'(x)\) are identically zero.

Proof.
Fix \(a \in \R\text{.}\) Define
\begin{equation*} F(x)=\sin(x+a) \qquad\text{and}\qquad G(x)=\sin x \cos a+\cos x \sin a. \end{equation*}
By Proposition 8.4.2, both functions belong to \(C^2(\R)\) and satisfy \(y''=-y\text{.}\) Also,
\begin{equation*} F(0)=\sin a=G(0) \qquad\text{and}\qquad F'(0)=\cos a=G'(0). \end{equation*}
Therefore \(H=F-G\) satisfies the hypotheses of Lemma 8.4.4, so \(H=0\text{.}\) This proves the sine addition formula.
For cosine, define
\begin{equation*} \widetilde{F}(x)=\cos(x+a) \qquad\text{and}\qquad \widetilde{G}(x)=\cos x \cos a-\sin x \sin a. \end{equation*}
Again both satisfy \(y''=-y\text{,}\) and
\begin{equation*} \widetilde{F}(0)=\cos a=\widetilde{G}(0), \qquad \widetilde{F}'(0)=-\sin a=\widetilde{G}'(0). \end{equation*}
Hence \(\widetilde{F}=\widetilde{G}\) by the same lemma.

Proof.
For \(0 \lt x \le 2\text{,}\) the successive term magnitudes in the sine series satisfy
\begin{equation*} \frac{x^{2n+3}/(2n+3)!}{x^{2n+1}/(2n+1)!} = \frac{x^2}{(2n+2)(2n+3)} \le \frac{4}{6} \lt 1. \end{equation*}
Thus the terms decrease in absolute value to \(0\text{,}\) so the alternating-series test gives
\begin{equation*} 0 \lt \sin x \le x. \end{equation*}

Proof.
By Proposition 8.4.2 and Proposition 8.4.6, we have
\begin{equation*} (\cos x)'=-\sin x \lt 0 \qquad \text{for } 0 \lt x \le 2. \end{equation*}
Hence \(\cos x\) is strictly decreasing on \([0,2]\text{.}\)
Using the alternating-series estimate,
\begin{equation*} \cos 1 = 1-\frac{1}{2!}+\frac{1}{4!}-\cdots \gt 1-\frac12 \gt 0, \end{equation*}
while
\begin{equation*} \cos 2 = 1-\frac{2^2}{2!}+\frac{2^4}{4!}-\frac{2^6}{6!}+\cdots \lt 1-2+\frac{16}{24} = -\frac13 \lt 0. \end{equation*}
Therefore the Intermediate Value Theorem gives a zero of \(\cos x\) in \((1,2)\text{,}\) and strict decrease makes that zero unique.
We define \(\pi\) to be twice this number:
\begin{equation*} \pi := 2(\pi/2). \end{equation*}

Proof.
Since \(\cos(\pi/2)=0\) and \(\sin^2(\pi/2)+\cos^2(\pi/2)=1\text{,}\) we have \(\sin(\pi/2)=1\text{.}\) The addition formulas with \(a=\pi/2\) therefore give
\begin{equation*} \begin{aligned} \sin\left(x+\frac{\pi}{2}\right) &= \sin x \cos\left(\frac{\pi}{2}\right) + \cos x \sin\left(\frac{\pi}{2}\right)\\ &= \cos x, \end{aligned} \end{equation*}
and
\begin{equation*} \begin{aligned} \cos\left(x+\frac{\pi}{2}\right) &= \cos x \cos\left(\frac{\pi}{2}\right) - \sin x \sin\left(\frac{\pi}{2}\right)\\ &= -\sin x. \end{aligned} \end{equation*}
Putting \(x=\pi/2\) into the first identity gives \(\sin(\pi)=0\text{,}\) and then the second identity yields \(\cos(\pi)=-1\text{.}\) Applying the addition formulas with \(a=\pi\) gives
\begin{equation*} \sin(x+\pi)=\sin x \cos \pi+\cos x \sin \pi=-\sin x, \end{equation*}
and
\begin{equation*} \cos(x+\pi)=\cos x \cos \pi-\sin x \sin \pi=-\cos x. \end{equation*}
Replacing \(x\) by \(x+\pi\) in these formulas shows that
\begin{equation*} \sin(x+2\pi)=\sin x, \qquad \cos(x+2\pi)=\cos x. \end{equation*}

Proof.
By Proposition 8.4.3, we have \(|\cos x| \le 1\) for all \(x\text{.}\) Applying the Mean Value Theorem to \(\sin\) on the interval between \(0\) and \(x\text{,}\) we obtain
\begin{equation*} |\sin x-\sin 0| = |\cos c|\,|x| \end{equation*}
for some \(c\) between \(0\) and \(x\text{.}\) Since \(\sin 0=0\text{,}\) it follows that \(|\sin x| \le |x|\text{.}\)
Dividing the power series for \(\sin x\) by \(x\) gives
\begin{equation*} \frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots \qquad (x \neq 0). \end{equation*}
The right-hand side converges to \(1\) as \(x \to 0\text{,}\) and the power series for \(\cos x\) plainly tends to \(1\) as well.
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