化解[cos(x/2)+sin(x/2)]*[cos(x/2)-sin(x/2)]*[1+tanx*tan(x/2)]和[2(cosx)^4-2(cosx)^2+1/2]/{2tan(π/4-x)*[sin(π/4+x)]^2}和(sinx)^2*(siny)^2+(cosx)^2*(cosy)^2-(1/2)*cos(2x)*cos(2y)

Question

化解[cos (x/2) + sin (x/2)]* [cos (x/2) - sin (x/2)]* [1+tan x * tan( x/2)]
和[2(cos x)^4 -2(cos x)^2 + 1/2] / {2tan(π/4 -x)* [sin(π/4 +x)]^2}
和(sin x)^2 *(sin y)^2+(cosx)^2 *(cosy)^2 -(1/2)*cos(2x)*cos(2y)

Answer 1

[cos (x/2) + sin (x/2)]* [cos (x/2) - sin (x/2)]* [1+tan x * tan( x/2)]
=cosx[1+tan x * tan( x/2)]
=cosx+sinxtanx/2=cosx+2(sinx/2)²=1
[2(cos x)^4 -2(cos x)^2 + 1/2] / {2tan(π/4 -x)* [sin(π/4 +x)]^2}
=0.5（2cos²x-1）²/[2cot(π/4+x)sin²(π/4+x)]
=0.5cos²2x/[2cos(π/4+x)sin(π/4+x)]
=0.5cos²2x/sin(π/2+2x)=0.5cos²2x/cos2x=0.5cos2x
(sin x)^2 *(sin y)^2+(cosx)^2 *(cosy)^2 -(1/2)*cos(2x)*cos(2y)
=(sin x)^2 *(sin y)^2+(cosx)^2 *(cosy)^2 -(1/2)*cos(2x)*cos(2y)+2sinxsinycosxcosy-2sinxsinycosxcosy
=(sin x)^2 *(sin y)^2-2sinxsinycosxcosy+(cosx)^2 *(cosy)^2-(1/2)*cos(2x)*cos(2y)+2sinxsinycosxcosy
=(sinxsiny-cosxcosy)²+（1/2）sin(2x)*sin(2y)-(1/2)*cos(2x)*cos(2y)
=cos²(x+y)-0.5cos(2x+2y)
=1/2