CoS(2Ⅹ(2α))=1-2Sⅰn^2α是怎么得到的有详解说明吗？
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cos(A+B)= cosA.cosB -sinA.sinB
A=B=α
cos(2α)
=(cosα)^2 -(sinα)^2
=[1-(sinα)^2] -(sinα)^2
=1-2(sinα)^2
A=B=α
cos(2α)
=(cosα)^2 -(sinα)^2
=[1-(sinα)^2] -(sinα)^2
=1-2(sinα)^2
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cos(2α) = cos(α+α) = cosαcosα-sinαsinα
= (cosα)^2 - (sinα)^2 = 1- (sinα)^2 - (sinα)^2
= 1-2(sinα)^2
cos(4α) = cos(2α+2α) = cos2αcos2α-sin2αsin2α
= (cos2α)^2 - (sin2α)^2 = 1- (sin2α)^2 - (sin2α)^2
= 1-2(sin2α)^2
= (cosα)^2 - (sinα)^2 = 1- (sinα)^2 - (sinα)^2
= 1-2(sinα)^2
cos(4α) = cos(2α+2α) = cos2αcos2α-sin2αsin2α
= (cos2α)^2 - (sin2α)^2 = 1- (sin2α)^2 - (sin2α)^2
= 1-2(sin2α)^2
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你的回答书上也有，关件是COs4α不是CoS2α难到cos4a与CoS2a都是一样的结果吗′
追答
cos4α = 1 - 2(sin2α)^2。 你写的式子不对，自然推不出。
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