化简[2cos^4(x)-2cos^2(x)+1/2]/[2tan(π/4-x)*sin^2(π/4+x)]
问题描述:
化简[2cos^4(x)-2cos^2(x)+1/2]/[2tan(π/4-x)*sin^2(π/4+x)]
答
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