Exercice 2 --- (id : 828)
Trigonométrie: Exercice 2
correction
1
a Dans le cercle circonscrit au triangle ABCABC (de diamètre [BC]) AOC^\widehat{AOC} est l'angle au centre associé à l'angle inscrit ABC^\widehat{ABC} donc AOH^=AOC^=2×ABC^=2×π12=π6\widehat{AOH}=\widehat{AOC}=2\times\widehat{ABC}=2\times\dfrac{\pi}{12}=\dfrac{\pi}{6}
Dans le triangle AOHAOH, A^+O^+H^=π\widehat{A}+\widehat{O}+\widehat{H}=\pi donc A^=πO^H^=ππ6π2=π3\widehat{A}=\pi-\widehat{O}-\widehat{H}=\pi-\dfrac{\pi}{6}-\dfrac{\pi}{2}=\dfrac{\pi}{3}
b 🔹 cosO^=OHOA\cos\widehat{O}=\dfrac{OH}{OA}     OH=OA×cosO^\iff OH=OA\times\cos\widehat{O}     OH=BC2×cos(π6)\iff OH=\dfrac{BC}{2}\times\cos\left({\dfrac{\pi}{6}}\right)     OH=4×32=23\iff OH=4\times \dfrac{\sqrt{3}}{2}=2\sqrt{3}
🔹 Le triangle AOHAOH est rectangle en HH donc AH2=OA2OH2AH^2=OA^2-OH^2     AH=OA2OH2\iff AH=\sqrt{OA^2-OH^2}     AH=1612=2\iff AH=\sqrt{16-12}=2
🔹 Dans le triangle AHBAHB rectangle en HH on a: HB=HO+OB=23+4HB=HO+OB=2\sqrt{3}+4
AB2=HA2+HB2AB^2=HA^2+HB^2
    AB2=4+(23+4)2\iff AB^2=4+(2\sqrt{3}+4)^2
    AB2=4+12+16+163\iff AB^2=4+12+16+16\sqrt{3}
    AB2=32+163=16(2+3)\iff AB^2=32+16\sqrt{3}=16(2+\sqrt{3})
    AB=42+3\iff AB=4\sqrt{2+\sqrt{3}}
2
a Le triangle AHBAHB est rectangle en HH donc :
🔹 cosB^=BHBA\cos\widehat{B}=\dfrac{BH}{BA}     cosB^=23+442+3\iff \cos\widehat{B}=\dfrac{2\sqrt{3}+4}{4\sqrt{2+\sqrt{3}}}     cos(π12)=2+32\iff \boxed{\cos\left({\dfrac{\pi}{12}}\right)=\dfrac{\sqrt{2+\sqrt{3}}}{2}}
🔹 sinB^=AHAB\sin\widehat{B}=\dfrac{AH}{AB}     sin(π12)=242+3\iff \sin\left({\dfrac{\pi}{12}}\right)=\dfrac{2}{4\sqrt{2+\sqrt{3}}}     sin(π12)=232(2+3)(23)\iff \sin\left({\dfrac{\pi}{12}}\right)=\dfrac{\sqrt{2-\sqrt{3}}}{2\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}     sin(π12)=232\iff \boxed{\sin\left({\dfrac{\pi}{12}}\right)=\dfrac{\sqrt{2-\sqrt{3}}}{2}}
b 🔹 cos5π12=cos(π2π12)\cos\dfrac{5\pi}{12}=\cos\left({\dfrac{\pi}{2}-\dfrac{\pi}{12}}\right) =sinπ12=232=\sin\dfrac{\pi}{12}=\dfrac{\sqrt{2-\sqrt{3}}}{2}
🔹 sin5π12=sin(π2π12)\sin\dfrac{5\pi}{12}=\sin\left({\dfrac{\pi}{2}-\dfrac{\pi}{12}}\right) =cosπ12=2+32=\cos\dfrac{\pi}{12}=\dfrac{\sqrt{2+\sqrt{3}}}{2}
🔹 cos11π12=cos(ππ12)\cos\dfrac{11\pi}{12}=\cos\left({\pi-\dfrac{\pi}{12}}\right) =cosπ12=2+32=-\cos\dfrac{\pi}{12}=-\dfrac{\sqrt{2+\sqrt{3}}}{2}
🔹 sin11π12=sin(ππ12)\sin\dfrac{11\pi}{12}=\sin\left({\pi-\dfrac{\pi}{12}}\right) =sinπ12=232=\sin\dfrac{\pi}{12}=\dfrac{\sqrt{2-\sqrt{3}}}{2}
solution de l'exercice n°2