Question

Вычислить 841sin2α, если известно, что cos(3π/4+α)=− 41√2/58

Answer 1

$\cos(3\pi/4+\alpha )=-\frac{41\sqrt{2} }{58} \\\cos(3\pi/4)\cos(\alpha)-\sin(3\pi/4)\sin(\alpha)=-\frac{41\sqrt{2} }{58} \\-\frac{\sqrt{2} }{2}\cos(\alpha)-\frac{\sqrt{2} }{2}\sin(\alpha)=-\frac{41\sqrt{2} }{58} \\-\frac{\sqrt{2} }{2}(\cos(\alpha)+\sin(\alpha))=-\frac{41\sqrt{2} }{58}\\\cos(\alpha)+\sin(\alpha)=\frac{41}{29}\\(\cos(\alpha)+\sin(\alpha))^2=(\frac{41}{29})^2\\\cos^2(\alpha)+2\sin(\alpha)\cos(\alpha)+\sin^2(\alpha)=\frac{1681}{841}\\$

$1+\sin(2\alpha)=\frac{1681}{841}\\\sin(2\alpha)=\frac{1681}{841}-1\\\sin(2\alpha)=\frac{840}{841}\\841\sin(2\alpha)=840$