Question

sin(2arcsin(-1/4))


помогите пожалуйста​

Answer 1

$\qquad \boxed {sin2\alpha =2\cdot sin\alpha \cdot cos\alpha }\\\\\\sin(2\, arcsin(-\frac{1}{4}))=2\cdot sin(arcsin(-\frac{1}{4}))\cdot cos(arcsin(-\frac{1}{4}))=\\\\=2\cdot (-\frac{1}{4})\cdot cos(-arcsin\frac{1}{4})=-\frac{1}{2}\cdot cos(arcsin\frac{1}{4})=\\\\\Big [\; cos(arcsin\frac{1}{4})=cos\alpha \; ;\; \; \alpha =arcsin\frac{1}{4}\; \Rightarrow \; sina=\frac{1}{4}>0\; ,\; \; -\frac{\pi}{2}\leq \alpha \leq \frac{\pi}{2}\; ,\\\\cosa=+\sqrt{1-sin^2\alpha }=\sqrt{1-(\frac{1}{4})^2}=\frac{\sqrt{15}}{4}\; \Big ]=$

$=-\frac{1}{2}\cdot \frac{\sqrt{15}}{4}=-\frac{\sqrt{15}}{8}$