« Trigonométrie/Exercices/Relations trigonométriques 2 » : différence entre les versions
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→Exercice 11-6 : sol |
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'''3°''' <math>\cos(a+b)\cos(a-b)+\cos(b+c)\cos(b-c)+\cos(c+a)\cos(c-a)=2(\cos^2a+\cos^2b+\cos^2c)-3</math>.
{{Solution|contenu=
'''1°''' <math>\sin(a+b)\sin(a-b)+\sin(b+c)\sin(b-c)+\sin(c+a)\sin(c-a)=\frac{\cos2b-\cos2a+\cos2c-\cos2b+\cos2a-\cos2c}2=0</math>.
Ligne 81 ⟶ 83 :
'''3°''' <math>\cos(a+b)\cos(a-b)+\cos(b+c)\cos(b-c)+\cos(c+a)\cos(c-a)=\cos2a+\cos2b+\cos2c=2(\cos^2a+\cos^2b+\cos^2c)-3</math>.
'''4°''' <math>\left(\tan a+\tan b+\tan c-\tan a\tan b\tan c\right)\cos a\cos b\cos c=\sin a\cos b\cos c+\sin b\cos c\cos a+\sin c\cos a\cos b-\sin a\sin b\sin c=\sin(a+b)\cos c+\cos(a+b)\sin c=\sin(a+b+c)</math>.▼
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== Exercice 11-6 ==
▲:<math>\tan a+\tan b+\tan c-\tan a\tan b\tan c=\frac{\sin(a+b+c)}{\cos a\cos b\cos c}</math>.
▲<math>\left(\tan a+\tan b+\tan c-\tan a\tan b\tan c\right)\cos a\cos b\cos c=\sin a\cos b\cos c+\sin b\cos c\cos a+\sin c\cos a\cos b-\sin a\sin b\sin c=\sin(a+b)\cos c+\cos(a+b)\sin c=\sin(a+b+c)</math>.
Démontrez les identités :
'''1°''' <math>\left(\cot
'''2°''' <math>\cot\left(\frac
'''1°''' Soit <math>t=\tan a</math>.▼
:<math>2\,\frac{3+\cos4a}{1-\cos4a}=\frac{2+2\cos^22a}{\sin^22a}=2\frac{1+\left(\frac{1-t^2}{1+t^2}\right)^2}{\left(\frac{2t}{1+t^2}\right)^2}=\frac{\left(1+t^2\right)^2+\left(1-t^2\right)^2}{2t^2}=\frac1{t^2}+t^2</math>.▼
'''
'''
{{Solution|contenu=
'''1°''' <math>\frac4{1-2\tan a\cot2a}=\frac4{\tan^2t}=\left(\frac{1-\tan^2\frac a2}{\tan\frac a2}\right)^2=\left(\cot\frac a2-\tan\frac a2\right)^2</math>.
'''2°''' <math>\cot b+\tan b=\frac{1+\tan^2b}{\tan b}=\frac2{\sin2b}=\frac2{\cos\left(2b-\frac\pi2\right)}</math>.
▲:<math>2\,\frac{3+\cos4a}{1-\cos4a}=\frac{2+2\cos^22a}{\sin^22a}=2\frac{1+\left(\frac{1-t^2}{1+t^2}\right)^2}{\left(\frac{2t}{1+t^2}\right)^2}=\frac{\left(1+t^2\right)^2+\left(1-t^2\right)^2}{2t^2}=\frac1{t^2}+t^2</math>.
'''4°''' <math>\frac{\sin2a}{1+\cos2a}\times\frac{\cos a}{1+\cos a}=\frac{2\sin a\cancel{\cos^2a}}{4\cancel{\cos^2a}\cos^2\frac a2}=\frac{\cancel4\cos\frac a2\sin\frac a2}{\cancel4\cos^2\frac a2}=\tan\frac a2</math>.
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