Question

Вычислить:
36 √2 sin2a, если tg a=-2 √2
90’

Answer 1

$\displaystyle\bf\\90^\circ < \alpha < 180^\circ \ \ \ \Rightarrow \ \ \ Cos\alpha < 0\\\\1+tg^{2}\alpha =\frac{1}{Cos^{2} \alpha } \\\\\\Cos^{2} \alpha =\frac{1}{1+tg^{2} \alpha } =\frac{1}{1+(-2\sqrt{2} )^{2} } =\frac{1}{1+8} =\frac{1}{9} \\\\\\Cos\alpha =-\sqrt{\frac{1}{9} } =-\frac{1}{3} \\\\\\tg\alpha =\frac{Sin\alpha }{Cos\alpha }\\\\\\Sin\alpha =tg\alpha \cdot Cos\alpha =-2\sqrt{2} \cdot\Big(-\frac{1}{3} \Big)=\frac{2\sqrt{2} }{3}$

$\displaystyle\bf\\Sin2\alpha =2Sin\alpha Cos\alpha =2\cdot \frac{2\sqrt{2} }{3} \cdot\Big(-\frac{1}{3} \Big)=-\frac{4\sqrt{2} }{9} \\\\\\36\sqrt{2} \ Sin2\alpha =36\sqrt{2} \cdot \Big(-\frac{4\sqrt{2} }{9} \Big)=-32\\\\\\Otvet \ : \ 36\sqrt{2} \ Sin2\alpha =-32$