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化简:2cos4X+1/2×sin22X-cos2X
 
人气:326 ℃ 时间:2020-06-26 22:44:39
解答
2cos4X+1/2×sin22X-cos2X
=2*[2(cos2x)^2-1]+1/2*(sin2x)^2-cos2x
=4(cos2x)^2-2+1/2*[1-(cos2x)^2]-cos2x
=7(cos2x)^2/2-cos2x-3/2
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