Trigonométrie : Exercice 3 2ème année secondaire

1 Pour tout réel x on a : $$\begin{align*} &(\cos x+\sin x)^2+(\cos x -\sin x)^2 \\ &=\underbrace{\cos^2x+\sin^2x}_{=1}+\cancel{2\cos x\sin x}+\underbrace{\cos^2x+\sin^2x}_{=1}-\cancel{2\cos x\sin x}\\ &=1+1\\ &=2 \end{align*}$$

2 Pour $x\in \left[{0;\pi}\right]$, $\sin x\geqslant 0$
$\cos x-\sin x=\dfrac{1}{2}\geqslant 0$ $\Longrightarrow \cos x\geqslant \sin x\geqslant 0$ $\Longrightarrow \cos x\geqslant 0$
$(\cos x+\sin x)^2+(\cos x -\sin x)^2=2$
$\iff (\cos x+\sin x)^2+\left({\dfrac{1}{2}}\right)^2=2$
$\iff (\cos x+\sin x)^2+\dfrac{1}{4}=2$
$\iff (\cos x+\sin x)^2=2-\dfrac{1}{4}=\dfrac{7}{4}$
$\iff \cos x+\sin x=\sqrt{\dfrac{7}{4}}$ car $\cos x+\sin x\geqslant0$
$\left\{{\begin{aligned}&{\cos x+\sin x=\dfrac{\sqrt{7}}{2}}\\&{\cos x-\sin x=\dfrac{1}{2}}\end{aligned}}\right.$ $\iff \left\{{\begin{aligned}&{2\cos x=\dfrac{\sqrt{7}+1}{2}}\\&{2\sin x=\dfrac{\sqrt{7}-1}{2}}\end{aligned}}\right.$ $\iff \left\{{\begin{aligned}&{\cos x=\dfrac{\sqrt{7}+1}{4}}\\&{\sin x=\dfrac{\sqrt{7}-1}{4}}\end{aligned}}\right.$