Exercice 4 --- (id : 809)
Trigonométrie: Exercice 4
correction
Pour x]0,π[x\in \left]{0,\pi}\right[; on pose f(x)=11+cosx+11cosxf(x)=\dfrac{1}{1+\cos x}+\dfrac{1}{1-\cos x}
1
  • f(π2)=11+0+110f\left({\dfrac{\pi}{2}}\right)=\dfrac{1}{1+0}+\dfrac{1}{1-0} =2=2
  • f(π3)=11+cos(π3)+11cos(π3)f\left({\dfrac{\pi}{3}}\right)=\dfrac{1}{1+\cos\left({\dfrac{\pi}{3}}\right)}+\dfrac{1}{1-\cos\left({\dfrac{\pi}{3}}\right)} =11+12+1112=\dfrac{1}{1+\dfrac{1}{2}}+\dfrac{1}{1-\dfrac{1}{2}} =23+2=83=\dfrac{2}{3}+2=\dfrac{8}{3}
  • f(π6)f\left({\dfrac{\pi}{6}}\right)=11+cos(π6)+11cos(π6)\dfrac{1}{1+\cos\left({\dfrac{\pi}{6}}\right)}+\dfrac{1}{1-\cos\left({\dfrac{\pi}{6}}\right)} =11+32+1132=\dfrac{1}{1+\dfrac{\sqrt{3}}{2}}+\dfrac{1}{1-\dfrac{\sqrt{3}}{2}} =22+3+223=\dfrac{2}{2+\sqrt{3}}+\dfrac{2}{2-\sqrt{3}} =2(23+2+3(2+3)(23))=2\left({\dfrac{2-\sqrt{3}+2+\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}}\right) =2(443)=8=2\left({\dfrac{4}{4-3}}\right)=8
2 f(x)=11+cosx+11cosx=1cosx(1+cosx)(1cosx)+1+cosx(1+cosx)(1cosx)=1cosx+1+cosx1cos2x=2sin2x\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*}
3 f(x)=4f(x)=4     2sin2x=4\iff \dfrac{2}{\sin^2x}=4     sin2x=24\iff \sin^2x=\dfrac{2}{4}     sinx=22\iff \sin x=\dfrac{\sqrt{2}}{2} ou sinx=22\sin x=-\dfrac{\sqrt{2}}{2} (à rejeter car x]0;π[sinx>0x\in \left]{0;\pi}\right[\Longrightarrow \sin x>0)     x=π4\iff x=\dfrac{\pi}{4} ou x=ππ4=3π4x=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}