Question

Вычислить sin2α и соs2α, если sinα= 1/18 и П/2 < a < П

Answer 1

$\displaystyle\bf\\\frac{\pi }{2} < \alpha < \pi \ \ \ \Rightarrow \ \ \ Cos\alpha < 0\\\\\\Sin\alpha =\frac{1}{18} \\\\\\Cos\alpha =-\sqrt{1-Sin^{2} \alpha }=-\sqrt{1-\Big(\frac{1}{18} \Big)^{2} } =-\sqrt{1-\frac{1}{324} } =-\sqrt{\frac{323}{324} } =-\frac{\sqrt{323} }{18} \\\\\\Sin2\alpha =2Sin\alpha Cos\alpha =2\cdot\frac{1}{18}\cdot\Big(-\frac{\sqrt{323} }{18}\Big)=-\frac{\sqrt{323} }{162}$

$\displaystyle\bf\\Cos2\alpha =1-2Sin^{2} \alpha =1-2\cdot \Big(\frac{1}{18} \Big)^{2} =1-2\cdot \frac{1}{324} =1-\frac{1}{162} =\frac{161}{162}$