Réponses

2014-01-28T11:30:12+01:00
Bonjour,

Si α ∈ ]π ; 3π/2[, alors sin α < 0 , cos α < 0 et tan α >0.

cos^2\alpha+sin^2\alpha=1\\\\cos^2\alpha=1-sin^2\alpha\\\\cos^2\alpha=1-(-\dfrac{3}{5})^2\\\\cos^2\alpha=1-\dfrac{9}{25}\\\\cos^2\alpha=\dfrac{25}{25}-\dfrac{9}{25}\\\\cos^2\alpha=\dfrac{16}{25}\\\\cos\ \alpha=-\dfrac{4}{5}\ \ ou\ \  cos\ \alpha=\dfrac{4}{5}

or  cos α < 0.

Donc  cos\alpha=-\dfrac{4}{5}


tan\ \alpha=\dfrac{sin\ \alpha}{cos\ \alpha}\\\\\\tan\ \alpha=\dfrac{-\dfrac{3}{5}}{-\dfrac{4}{5}}\\\\\\tan\ \alpha=-\dfrac{3}{5}\times(-\dfrac{5}{4})\\\\\\tan\ \alpha=\dfrac{3}{4}