新高一课改数学题

解：根号2*cosx-根号6*sinx
=2*(根号2)*[1/2*cosx-(根号3)/2*sinx]
=2*(根号2)*[cos(兀/3)*cosx-sin(兀/3)*sinx]
=2*(根号2)*cos(兀/3+x)

解：sin(a-b)*cosa-cos(b-a)*sina=3/5
sin(a-b)*cosa-cos(a-b)*sina=3/5
sin(a-b-a)=3/5
sin(-b)=3/5
sinb=-3/5
因为pi<b<3/2*pi
所以cosb=-4/5
所以sin(b+1/4*pi)
=sinb*cos(pi/4)+cosb*sin(pi/4)
=(-3/5)*(根号2)/2+(-4/5)*(根号2)/2
=(根号2)/2*(-3/5-4/5)
=(根号2)/2*(-7/5)
=-7/10*(根号2)
因为sin(b+5/4*pi)
=sin(pi+b+pi/4)
=-sin(b+pi/4)
=7/10*根号2

1.2根号2sin(π/6-x)
过程：根号2cosx-根号6sinx =2根号2(1/2cosx-根号3/2sinx)=2根号2sin(π/6-x)
2.7根号2/10
过程：由已知式得 sin(-b)=3/5,sinb=-3/5,
b是第三象限的角，故 cosb=-4/5,于是
sin（b+5π/4)=sinbcos(5π/4)+cosbsin(5π/4)
=-3/5*(-根号2/2)-4/5*(-根号2/2)=7根号2/10