Exercice 3 --- (id : 836)
Trigonométrie: Exercice 3
correction
1
a F(x)=cos3xsin2x14cosx+34=cos3x(1cos2x)14cosx+34=cos3x+cos2x14cosx14=cos2x(cosx+1)14(cosx+1)=(cos2x14)(cosx+1)\begin{align*} F(x)&=\cos^3x-\sin^2x-\dfrac{1}{4}\cos x+\dfrac{3}{4}\\ &=\cos^3x-(1-\cos^2x)-\dfrac{1}{4}\cos x+\dfrac{3}{4}\\ &=\cos^3x+\cos^2x-\dfrac{1}{4}\cos x-\dfrac{1}{4}\\ &=\cos^2x(\cos x+1)-\dfrac{1}{4}(\cos x+1)\\ &=(\cos^2x-\dfrac{1}{4})(\cos x+1) \end{align*}
b Soit x[0,π]x\in [0,\pi] tel que : F(x)=0    (cos2x14)(cosx+1)=0    cos2x14=0  ou  cosx+1=0    cos2x)14=0  ou  cosx=1    (cosx12)(cosx+12)=0  ou  x=π    cosx12=0  ou  cosx+12=0  ou  x=π    cosx=12  ou  cosx=12  ou  x=π    x=π3  ou  x=2π3  ou  x=πDonc      S[0;π]={π3,2π3,π}\begin{align*} &F(x)=0\\ &\iff (\cos^2x-\dfrac{1}{4})(\cos x+1)=0\\ &\iff \cos^2x-\dfrac{1}{4}=0\;ou\;\cos x+1=0\\ &\iff \cos^2x)-\dfrac{1}{4}=0\; ou \;\cos x=-1\\ &\iff (\cos x-\dfrac{1}{2})(\cos x+\dfrac{1}{2})=0\; ou \;x=\pi\\ &\iff \cos x-\dfrac{1}{2}=0\;ou\;\cos x+\dfrac{1}{2}=0\;ou\;x=\pi\\ &\iff \cos x=\dfrac{1}{2}\;ou\;\cos x=-\dfrac{1}{2}\;ou\;x=\pi\\ &\iff x=\dfrac{\pi}{3}\;ou\;x=\dfrac{2\pi}{3}\;ou\;x=\pi\\ &Donc\;\;\;\boxed{S_{[0;\pi]}=\left\{{\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi}\right\}} \end{align*}
2 A=cos2π8+cos23π8+cos25π8+cos27π8=cos2π8+cos23π8+cos2(π3π8)+cos2(ππ8)=cos2π8+cos23π8+(cos(3π8))2+(cos(π8))2=2(cos2π8+cos23π8)=2(cos2π8+sin2π8)    (car  π8+3π8=π2)=2×1=2\begin{align*} A&=\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{5\pi}{8}+\cos^2\dfrac{7\pi}{8}\\ &=\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\left({\pi-\dfrac{3\pi}{8}}\right)+\cos^2\left({\pi-\dfrac{\pi}{8}}\right)\\ &=\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\left({-\cos\left({\dfrac{3\pi}{8}}\right)}\right)^2+\left({-\cos\left({\dfrac{\pi}{8}}\right)}\right)^2\\ &=2\left({\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}}\right)\\ &=2\left({\cos^2\dfrac{\pi}{8}+\sin^2\dfrac{\pi}{8}}\right)\;\;(car\;\dfrac{\pi}{8}+\dfrac{3\pi}{8}=\dfrac{\pi}{2})\\ &=2\times 1=2 \end{align*}