化简:2cos4x−2 cos2x+ 1/2 2tan(π4−x)•sin2(π4+x) .
问题描述:
化简:
. 2cos4x−2 cos2x+
1 2 2tan(
−x)•sin2(π 4
+x) π 4
答
原式=
(4cos4x−4cos2x+1) 1 2 2•
•cos2(sin(
−x)π 4 cos(
−x)π 4
−x) π 4
=
(2cos2x−1)2
4sin(
−x)cos(π 4
−x)π 4
=
=cos22x 2sin(
−2x)π 2
=cos22x 2cos2x
cos2x.1 2