Question

найдите sin 2a и cos 2a если cos a 24/25
3п/2<а<2п​

Answer 1

α - угол четвёртой четверти, значит Sinα < 0 .

$Cos\alpha=\frac{24}{25}\\\\Sin\alpha=-\sqrt{1-Cos^{2}\alpha} =-\sqrt{1-(\frac{24}{25})^{2}}=-\sqrt{1-\frac{576}{625} }=-\sqrt{\frac{49}{625} }=-\frac{7}{25}\\\\Sin2\alpha=2Sin\alpha Cos\alpha=2*(-\frac{7}{25})*\frac{24}{25}=-\frac{336}{625}\\\\\boxed{Sin2\alpha=-\frac{336}{625}}\\\\\\Cos2\alpha =2Cos^{2}\alpha-1=2*(\frac{24}{25})^{2}-1=2*\frac{576}{625}-1=\frac{1152}{625}-1=\frac{527}{625}\\\\\boxed{Cos2\alpha=\frac{527}{625}}$

Answer 2

Объяснение:

$cos\alpha =\frac{24}{25} \ \ \ \ \frac{3\pi }{2} <\alpha <2\pi \ \ \ \ sin(2\alpha )=?\ \ \ \ cos(2\alpha )=?\\sin^2\alpha +cos^2\alpha =1\\sin^2=1-cos^2\alpha =1-(\frac{24}{25})^2=1-\frac{576}{625} =\frac{625-576}{625}=\frac{49}{625}.\\sin\alpha =б\sqrt{\frac{49}{625} }=б\frac{7}{25}. \\ \frac{3\pi }{2} <\alpha <2\pi\ \ \ \ \Rightarrow\\sin\alpha =-\frac{7}{25} .\\sin(2\alpha )=2*sin\alpha *cos\alpha =2*(-\frac{7}{25})*\frac{24}{25}=-\frac{336}{625} .$

$cos(2\alpha )=cos^2\alpha-sin^2\alpha =(\frac{24}{25})^2-(-\frac{7}{25})^2=\frac{576}{625}-\frac{49}{625}=\frac{527}{625}.$