Exercise 15.2.4

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Suppose that we define $\sin(x)$ by $y = \sin(x) \iff x = \int_0^y(1-t^2)^{-1/2} \D t$. Then define $\cos (x)$ to be $sin'(x)$. Use the method of Proposition 15.2.1 to prove the standard trigonometric identity $\cos^2(x) = 1 - \sin^2(x)$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
We obtain the analog of (15.9) as in Exercise 1: for all $x \in \R$,
\begin{align*}
\sin(-x) = - \sin(x),\
\sin(\pi-x) = \sin(x).
\end{align*}

Now we use the definition of $\sin$: for all $y\in [-1,1]$, for all $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,
$$y = \sin(x) \iff x = \int_0^y(1-t^2)^{-1/2} \D t,$$
where $\int_0^1(1-t^2)^{-1/2} \D t$ and $ \int_0^{-1}(1-t^2)^{-1/2} \D t$ converge.

If $x \in [0 , \frac{\pi}{2}[$, differentiating each side of
$$s = \int_0^{\sin(x)} \frac{1}{\sqrt{1-t^2}} \D t,$$
we obtain
$$1 = \frac{1}{\sqrt{1 - \sin^2(x)}} \sin'(x).$$
If $x = \frac{\pi}{2}$, then $\sin(x) = 1, \sin'(x) = 0$, thus $\sin'^2(x) = 1 - \sin^2(x)$.
Therefore
$$\cos(x) = \sin'(x) = \sqrt{1 - \sin^2(x)},\qquad 0 \leq x \leq \frac{\pi}{2}.$$
We extend the equality $\sin^2(x) + \cos^2(x) = 1$ to all $x \in \R$ as in Exercise 2.
\end{proof}

Exercise 15.2.4

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Exercise 15.2.4

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