三角函数化简

解:

由于:sin^3(x)sin3x

=sin^2(x)[sinx*sin3x]

=sin^2(x)[sinx*sin(2x+x)]

=sin^2(x)[sinx(sin2xcosx+sinxcos2x)]

=sin^2(x)[sinxcosx*sin2x+sin^2(x)cos2x]

=(1-cos2x)/2*[(1/2)(2sinxcosx)sin2x+(1-cos2x)/2*cos2x]

=[(1-cos2x)/4]*[sin^2(2x)+(1-cos2x)cos2x]

=[(1-cos2x)/4]*[1-cos^2(2x)+cos2x-cos^2(2x)]

=(1/4)(1-cos2x)*[-2cos^2(2x)+cos2x+1]

cos^3(x)cos3x

=cos^2(x)[cosx*cos(2x+x)]

=cos^2(x)[cosx(cos2xcosx-sin2xsinx)]

=cos^2(x)[cos^2(x)cos2x-sin2xsinxcosx]

=(1+cos2x)/2*[(1+cos2x)/2*cos2x-(1/2)(2sinxcosx)sin2x]

=(1+cos2x)/4*[(1+cos2x)cos2x-sin^2(2x)]

=(1/4)(1+cos2x)*[2cos^2(2x)+cos2x-1]

设cos2x=t

则：

[sin3xsin^3(x)+cos3xcos^3(x)]/cos^2(2x)

=[(1/4)(1-t)(-2t^2+t+1)+(1/4)(1+t)(2t^2+t-1)]/t^2

={(1/4)[(-2t^2+t+1+2t^3-t^2-t)+(2t^2+t-1+2t^3+t^2-t)]}/t^2

=[(1/4)(4t^3)]/t^2

=t^3/t^2

=t=cos2x

则：原式

=cos2x+sin2x

=√2sin(2x+∏/4)

(辅助角公式)

由于：sin(2x+∏/4)属于[-1，1]

则：√2sin(2x+∏/4)属于[-√2,√2]

其最小值为：-√2