trigo
Re: trigo
Bonjour
Exercice 1
1) $\gamma =\pi -(\alpha +\beta)$ donc $\sin \gamma =\sin (\alpha +\beta) =\sin \alpha \cos \beta +\sin \beta \cos \alpha$
$\sin \alpha +\sin \beta +\sin \gamma =\sin \alpha (\cos \beta +1) +\sin \beta (\cos \alpha +1)=\sin \alpha (2\cos^2 \frac{\beta}{2})+\sin \beta (2\cos^2 \frac{\alpha}{2})$
$=4\sin\frac{\alpha}{2}\cos \frac{\alpha}{2} \cos^2\frac{\beta}{2} +4\sin\frac{\beta}{2}\cos \frac{\beta}{2} \cos^2\frac{\alpha}{2}=4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}(\sin\frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\beta}{2} \cos \frac{\alpha}{2})$
$=4\cos \frac{\alpha}{2} \cos \frac{\beta}{2}\sin \frac{\alpha +\beta}{2}$
2) $\frac{\alpha +\beta}{2} =\frac{\pi -\gamma}{2}$ donc $\sin \frac{\alpha +\beta}{2} =\sin (\frac{\pi}{2} -\frac{\gamma}{2})=\cos \frac{\gamma}{2}$
Exercice 2
1) En réduisant au même dénominateur :
$\tan a +\tan b +\tan c =\frac{\sin a \cos b \cos c +\sin b \cos a \cos c +\sin c \cos a \cos b}{\cos a \cos b \cos c}$
$=\frac{\cos c (\sin a \cos b +\sin b \cos a ) +\sin c \cos a \cos b}{\cos a \cos b \cos c}=\frac{\cos c \sin (a+b)+\sin c \cos a \cos b}{\cos a \cos b \cos c}$
$\sin (a+b)=\sin (\pi -c)=\sin c$ et $\cos c =\cos (\pi -(a+b))=-\cos (a+b)=-\cos a \cos b +\sin a \sin b$
On obtient donc $\tan a +\tan b +\tan c=\frac{(-\cos a \cos b+\sin a \sin b)\sin c +\sin c \cos a \cos b}{\cos a \cos b \cos c}$
$=\frac{\sin a \sin b \sin c}{\cos a \cos b \cos c}=\tan a \tan b \tan c$
2) En remplaçant $\cos c$ par $-\cos (a+b)=-\cos a \cos b +\sin a \sin b$ et en réduisant, on obtient :
$\cos^2a +\cos^2 b -\cos^2 a \cos^2 b +\sin^2 a \sin^2 b$
$=\cos^2a (1-\cos^2 b) +\cos^2b +\sin^2a (1-\cos^2 b)=(1-\cos^2 b)(\cos^2a+\sin^2a)+\cos^2 b$
$=1-\cos^2 b +\cos^2 b =1$
Exercice 1
1) $\gamma =\pi -(\alpha +\beta)$ donc $\sin \gamma =\sin (\alpha +\beta) =\sin \alpha \cos \beta +\sin \beta \cos \alpha$
$\sin \alpha +\sin \beta +\sin \gamma =\sin \alpha (\cos \beta +1) +\sin \beta (\cos \alpha +1)=\sin \alpha (2\cos^2 \frac{\beta}{2})+\sin \beta (2\cos^2 \frac{\alpha}{2})$
$=4\sin\frac{\alpha}{2}\cos \frac{\alpha}{2} \cos^2\frac{\beta}{2} +4\sin\frac{\beta}{2}\cos \frac{\beta}{2} \cos^2\frac{\alpha}{2}=4\cos \frac{\alpha}{2}\cos \frac{\beta}{2}(\sin\frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\beta}{2} \cos \frac{\alpha}{2})$
$=4\cos \frac{\alpha}{2} \cos \frac{\beta}{2}\sin \frac{\alpha +\beta}{2}$
2) $\frac{\alpha +\beta}{2} =\frac{\pi -\gamma}{2}$ donc $\sin \frac{\alpha +\beta}{2} =\sin (\frac{\pi}{2} -\frac{\gamma}{2})=\cos \frac{\gamma}{2}$
Exercice 2
1) En réduisant au même dénominateur :
$\tan a +\tan b +\tan c =\frac{\sin a \cos b \cos c +\sin b \cos a \cos c +\sin c \cos a \cos b}{\cos a \cos b \cos c}$
$=\frac{\cos c (\sin a \cos b +\sin b \cos a ) +\sin c \cos a \cos b}{\cos a \cos b \cos c}=\frac{\cos c \sin (a+b)+\sin c \cos a \cos b}{\cos a \cos b \cos c}$
$\sin (a+b)=\sin (\pi -c)=\sin c$ et $\cos c =\cos (\pi -(a+b))=-\cos (a+b)=-\cos a \cos b +\sin a \sin b$
On obtient donc $\tan a +\tan b +\tan c=\frac{(-\cos a \cos b+\sin a \sin b)\sin c +\sin c \cos a \cos b}{\cos a \cos b \cos c}$
$=\frac{\sin a \sin b \sin c}{\cos a \cos b \cos c}=\tan a \tan b \tan c$
2) En remplaçant $\cos c$ par $-\cos (a+b)=-\cos a \cos b +\sin a \sin b$ et en réduisant, on obtient :
$\cos^2a +\cos^2 b -\cos^2 a \cos^2 b +\sin^2 a \sin^2 b$
$=\cos^2a (1-\cos^2 b) +\cos^2b +\sin^2a (1-\cos^2 b)=(1-\cos^2 b)(\cos^2a+\sin^2a)+\cos^2 b$
$=1-\cos^2 b +\cos^2 b =1$