Trigonométrie : Exercice 4 2ème année secondaire

Exercice 4 --- (id : 809)

correction

Pour

x\in \left]{0,\pi}\right[

; on pose

f(x)=\dfrac{1}{1+\cos x}+\dfrac{1}{1-\cos x}

$f\left({\dfrac{\pi}{2}}\right)=\dfrac{1}{1+0}+\dfrac{1}{1-0}$ $=2$
$f\left({\dfrac{\pi}{3}}\right)=\dfrac{1}{1+\cos\left({\dfrac{\pi}{3}}\right)}+\dfrac{1}{1-\cos\left({\dfrac{\pi}{3}}\right)}$ $=\dfrac{1}{1+\dfrac{1}{2}}+\dfrac{1}{1-\dfrac{1}{2}}$ $=\dfrac{2}{3}+2=\dfrac{8}{3}$
$f\left({\dfrac{\pi}{6}}\right)$ = $\dfrac{1}{1+\cos\left({\dfrac{\pi}{6}}\right)}+\dfrac{1}{1-\cos\left({\dfrac{\pi}{6}}\right)}$ $=\dfrac{1}{1+\dfrac{\sqrt{3}}{2}}+\dfrac{1}{1-\dfrac{\sqrt{3}}{2}}$ $=\dfrac{2}{2+\sqrt{3}}+\dfrac{2}{2-\sqrt{3}}$ $=2\left({\dfrac{2-\sqrt{3}+2+\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}}\right)$ $=2\left({\dfrac{4}{4-3}}\right)=8$

\begin{align*} f(x)&=\dfrac{1}{1+\cos x}+\dfrac{1}{1-\cos x}\\ &=\dfrac{1-\cos x}{(1+\cos x)(1-\cos x)}+\dfrac{1+\cos x}{(1+\cos x)(1-\cos x)}\\ &=\dfrac{1-\cos x+1+\cos x}{1-\cos^2x}\\ &=\dfrac{2}{\sin^2x} \end{align*}

f(x)=4

\iff \dfrac{2}{\sin^2x}=4

\iff \sin^2x=\dfrac{2}{4}

\iff \sin x=\dfrac{\sqrt{2}}{2}

\sin x=-\dfrac{\sqrt{2}}{2}

(à rejeter car

x\in \left]{0;\pi}\right[\Longrightarrow \sin x>0

)

\iff x=\dfrac{\pi}{4}

x=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}