Exercise 15.2.11

Proof. If $\alpha ,\beta \in [0,1]$, then there are unique $x,y\in [0,\pi ∕2]$ such that $\alpha =\sin <!--\; FUNCTION\; APPLICATION\; -->x$, $\beta =\sin <!--\; FUNCTION\; APPLICATION\; -->y$. Then $x=\arcsin (\alpha ),y=\arcsin (\beta ))$, where $\arcsin$ is the reciprocal function of $f$, $f$ being the restriction of $\sin <!--\; FUNCTION\; APPLICATION\; -->$ to $[0,\pi ]$. For every $t\in ]-1,1[$, $f$ is differentiable at $f^{-1}(t)\in ]0,\pi [$, and $f^{\prime }(f^{-1}(t)\ne 0$, thus $f^{-1}=\arcsin :[-1,1]\to [0,\pi ]$ is differentiable on $]-1,1[$, and for all $t\in ]-1,1[$,

Since $t\mapsto \frac{1}{\sqrt{1-t^2}}$ is continuous on $]-1,1[$, for all $x\in ]-1,1[$,

(This equality remains true for $x=\pm 1$:

${\int }_0^1\frac{1}{\sqrt{1-t^2}}dt$ is convergent, and ${\int }_0^1\frac{1}{\sqrt{1-t^2}}dt=\underset{x\to 1}{\lim <!--\; FUNCTION\; APPLICATION\; -->}{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}_0^x\frac{1}{\sqrt{1-t^2}}dt$, with value $\arcsin (1)=\pi ∕2$).

Therefore, for all $𝜃\in [0,\pi ]$, and for all $z\in [-1,1]$,

(Alternatively, we can take this equivalence as a definition of $\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃$, to continue Exercise 4.)

Write $\gamma =\sin <!--\; FUNCTION\; APPLICATION\; -->(x+y)$. Since $0\le x+y\le \pi$, we obtain $x+y=\arcsin (\gamma )$, that is

Moreover, since $x,y\in [0,\frac{\pi }{2}]$, $\cos <!--\; FUNCTION\; APPLICATION\; -->x\ge 0,\cos <!--\; FUNCTION\; APPLICATION\; -->y\ge 0$, thus

and