化简：2cos4x−2cos2x+122tan(π4−x)•sin2(π4+x)．

Question

化简：(2cos4x−2cos2x+(1)/(2))/(2tan((π)/(4)−x)•sin2((π)/(4)+x))．

Answer 1

原式=((1)/(2)(4cos4x−4cos2x+1))/(2•(sin((π)/(4)−x))/(cos((π)/(4)−x))•cos2((π)/(4)−x))
=((2cos2x−1)2)/(4sin((π)/(4)−x)cos((π)/(4)−x))
=(cos22x)/(2sin((π)/(2)−2x))=(cos22x)/(2cos2x)=(1)/(2)cos2x．