求证：cos2αcos2β=1/2{cos2（α+β）+cos2(α-β)}
1个回答

用a和b

左边=cos[(a+b)+(a-b)]cos[(a+b)-(a-b)]

=[cos(a+b)cos(a-b)-sin(a+b)sin(a-b)][cos(a+b)cos(a-b)+sin(a+b)sin(a-b)]

=cos²(a+b)cos²(a-b)-sin²(a+b)sin²(a-b)

=[1+cos2(a+b)]/2[1+cos2(a-b)]/2-[1-cos2(a+b)]/2[1-cos2(a-b)]/2

=[1+cos2(a+b)+cos2(a-b)+cos2(a+b)]/4-[1-cos2(a+b)-cos2(a-b)+cos2(a+b)]/4

=[cos2(a+b)+cos2(a-b)]/2=右边

命题得证

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