Trigonométrie

Exercices types : 11ère partie - Exercice 1

15 min
30
Question 1
On donne cos⁡(7π8)=−2+22\cos \left(\frac{7\pi }{8} \right)=-\sqrt{\frac{2+\sqrt{2} }{2} } et sin⁡(7π8)=2−22\sin \left(\frac{7\pi }{8} \right)=\sqrt{\frac{2-\sqrt{2} }{2} }

Calculer :
a.\bf{a.} π−7π8\pi -\frac{7\pi }{8}                                                                                                   \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; b.\bf{b.} π+7π8\pi +\frac{7\pi }{8}

c.\bf{c.} π2−7π8\frac{\pi }{2} -\frac{7\pi }{8}                                                                                                     \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; d.\bf{d.} π2+7π8\frac{\pi }{2} +\frac{7\pi }{8}

Correction
a.\red{\bf{a.}} π−7π8=π1−7π8=π×81×8−7π8=8π8−7π8=8π−7π8=π8\pi -\frac{7\pi }{8} =\frac{\pi }{1} -\frac{7\pi }{8} =\frac{\pi \times 8}{1\times 8} -\frac{7\pi }{8} =\frac{8\pi }{8} -\frac{7\pi }{8} =\frac{8\pi -7\pi }{8} ={\color{blue}{\frac{\pi }{8}}}
b.\red{\bf{b.}} π+7π8=π1+7π8=π×81×8+7π8=8π8+7π8=8π+7π8=15π8\pi +\frac{7\pi }{8} =\frac{\pi }{1} +\frac{7\pi }{8} =\frac{\pi \times 8}{1\times 8} +\frac{7\pi }{8} =\frac{8\pi }{8} +\frac{7\pi }{8} =\frac{8\pi +7\pi }{8} ={\color{blue}{\frac{15\pi }{8}}}
c.\red{\bf{c.}} π2−7π8=π×42×4−7π8=4π8−7π8=4π−7π8=−3π8\frac{\pi }{2} -\frac{7\pi }{8} =\frac{\pi \times 4}{2\times 4} -\frac{7\pi }{8} =\frac{4\pi }{8} -\frac{7\pi }{8} =\frac{4\pi -7\pi }{8} ={\color{blue}{-\frac{3\pi }{8}}}
d.\red{\bf{d.}} π2+7π8=π×42×4+7π8=4π8+7π8=4π+7π8=11π8\frac{\pi }{2} +\frac{7\pi }{8} =\frac{\pi \times 4}{2\times 4} +\frac{7\pi }{8} =\frac{4\pi }{8} +\frac{7\pi }{8} =\frac{4\pi +7\pi }{8} ={\color{blue}{\frac{11\pi }{8}}}
Question 2

En déduire :
a.\bf{a.} cos⁡(π8)\cos \left(\frac{\pi }{8} \right)                                                                                                   \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; b.\bf{b.} sin⁡(15π8)\sin \left(\frac{15\pi }{8} \right)

c.\bf{c.} sin⁡(−3π8)\sin \left(-\frac{3\pi }{8} \right)                                                                                         \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; d.\bf{d.} cos⁡(11π8)\cos \left(\frac{11\pi }{8} \right)

Correction
a.\red{\bf{a.}}
cos⁡(π−x)=−cos⁡(x)\cos\left(\pi-{\color{red}{x}}\right)=-\cos\left({\color{red}{x}}\right)
Ainsi :
cos⁡(π8)=cos⁡(π−7π8)\cos \left(\frac{\pi }{8} \right)=\cos\left(\pi-{\color{red}{\frac{7\pi}{8}}}\right)
cos⁡(π8)=−cos⁡(7π8)\cos \left(\frac{\pi }{8} \right)=-\cos\left({\color{red}{\frac{7\pi}{8}}}\right)
Finalement :\purple{\text{Finalement :}}
cos⁡(π8)=−(−2+22)=2+22\cos \left(\frac{\pi }{8} \right)=-\left(-\sqrt{\frac{2+\sqrt{2} }{2} }\right)=\sqrt{\frac{2+\sqrt{2} }{2} }

b.\red{\bf{b.}}
sin⁡(π+x)=−sin⁡(x)\sin\left(\pi+{\color{red}{x}}\right)=-\sin\left({\color{red}{x}}\right)
Ainsi :
sin⁡(15π8)=sin⁡(π+7π8)\sin\left(\frac{15\pi }{8}\right)=\sin\left(\pi+{\color{red}{\frac{7\pi}{8}}}\right)
sin⁡(15π8)=−sin⁡(7π8)\sin\left(\frac{15\pi }{8}\right)=-\sin\left({\color{red}{\frac{7\pi}{8}}}\right)
Finalement :\purple{\text{Finalement :}}
sin⁡(15π8)=−2−22\sin\left(\frac{15\pi }{8}\right)=-\sqrt{\frac{2-\sqrt{2} }{2} }

c.\red{\bf{c.}}
sin⁡(π2−x)=cos⁡(x)\sin\left(\frac{\pi }{2} -{\color{red}{x}}\right)=\cos\left({\color{red}{x}}\right)
Ainsi :
sin⁡(−3π8)=sin⁡(π2−7π8)\sin \left(-\frac{3\pi }{8} \right)=\sin\left(\frac{\pi }{2}-{\color{red}{\frac{7\pi}{8}}}\right)
sin⁡(−3π8)=cos⁡(7π8)\sin \left(-\frac{3\pi }{8} \right)=\cos\left({\color{red}{\frac{7\pi}{8}}}\right)
Finalement :\purple{\text{Finalement :}}
sin⁡(−3π8)=−2+22\sin \left(-\frac{3\pi }{8} \right)=-\sqrt{\frac{2+\sqrt{2} }{2} }

d.\red{\bf{d.}}
cos⁡(π2+x)=−sin⁡(x)\cos\left(\frac{\pi }{2} +{\color{red}{x}}\right)=-\sin\left({\color{red}{x}}\right)
Ainsi :
cos⁡(11π8)=cos⁡(π2+7π8)\cos \left(\frac{11\pi }{8} \right)=\cos\left(\frac{\pi }{2}+{\color{red}{\frac{7\pi}{8}}}\right)
cos⁡(11π8)=−sin⁡(7π8)\cos \left(\frac{11\pi }{8} \right)=-\sin\left({\color{red}{\frac{7\pi}{8}}}\right)
Finalement :\purple{\text{Finalement :}}
cos⁡(11π8)=−2−22\cos \left(\frac{11\pi }{8} \right)=-\sqrt{\frac{2-\sqrt{2} }{2} }