1.求 根号(1+2*sin610°*cos430°)/sin 250°+cos 790°

题目

1.求 根号(1+2*sin610°*cos430°)/sin 250°+cos 790°
PS:根号是1+2*sin610°*cos430°整个
2.已知tan a=-3,求下列各式的值:
(1)(4*sin a-2*cos a)/(5*cos a=3*sin a)
(2) (sin a-cos a)^2
要有具体过程

答案

1.√(1+2*sin610°*cos430°)/sin 250°+cos 790°
=√(1+2*sin(-110)°*cos70°)/sin110°+cos70°
=√(1-2*sin110°*cos70°)/sin110°+cos70°
令110°=x,70°=y
2*sin110°*cos70°=2*sinx*cosy=2*(1/2)*[sin(x+y)+sin(x-y)]=sin180°+sin40°=sin40°=sin(2*20°)=2*sin20°cos20°
√(1+2*sin610°*cos430°)=√1+2*sin20°cos20°=√（cos20+sin20）^2
=cos20°+sin20°
cos20°+sin20°=sin70°+sin20°
sin110°+cos70°=sin70°+sin20°
所以原式=（sin70°+sin20°)/(sin70°+sin20°)=1
2.tana=-3
所以 sina/cosa=-3
(1)(4*sin a-2*cos a)/(5*cos a=3*sin a)
则原式=(4(sina/cosa)-2)/(5-3(sina/cosa))
=(4*(-3)-2)/(5-3(-3))
=-14/14
=-1
(sin a-cos a)^2
=(sina)^2+(cosa)^2-2sina*cosa
=1-2sina*cosa
=1-sin2a
=1-[(2tana)/(1+(tana)^2)]
=1-(-6/10)
=8/5