化简:2cos^4x-2cos^2x+1/2 /2tan(π/4-x)sin^2(π/4+x)

Question

化简:2cos^4x-2cos^2x+1/2 /2tan(π/4-x)sin^2(π/4+x)

Answer 1

分母=2tan(π/4-x)*cos^2[π/2-(π/4+x)] =2[sin(π/4-x)/cos(π/4-x)]*cos^2(π/4-x) =2sin(π/4-x)cos(π/4-x) =sin[2(π/4-x)] =sin(π/2-2x) =cos2x 分子=1/2[4cos^4(x)-4cos^2(x)+1] =1/2(2cos^2x-1)^2 =1/2(cos2x)^2 所以原式=1/2(cos2x)^2/cos2x=1/2*cos2x