Question

化简cos(π/4-a)cos(π/4+a)

Answer 1

解法1：
套用公式：
cosα·cosβ=(1/2)[cos(α+β)+cos(α-β)]
有：
cos(π/4-a)cos(π/4+a)
=(1/2)[cos(π/4-a+π/4+a)cos(π/4-a-π/4-a)]
=(1/2)[cos(π/2)+cos(-2a)]
=(1/2)[0+cos(2a)]
=(1/2)cos(2a)

解法2：
cos(π/4-a)cos(π/4+a)
=[cos(π/4)cosa+sin(π/4)sina][cos(π/4)cosa-sin(π/4)sina]
={[(√2)/2]cosa+[(√2)/2]sina}{[(√2)/2]cosa-[(√2)/2]sina}
=(1/2)(cosa+sina)(cosa-sina)
=(1/2)[(cosa)^2-(sina)^2]
=(1/2)cos(2a)

Answer 2

该式：cos(a+b)*cos(a-b)=cos²a-sin²b

该题：
原式=cos²π/4-sin²a
=1/2-sin²a
=(1/2)(1-2sin²a)
=(1/2)cos2a

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Answer 3

解:
cos(π/4-a)cos(π/4+a)=cos(a-π/4)cos(a+π/4)
=cos(a-(π/2-π/4))cos(a+π/4)
=cos((a+π/4)-π/2)cos(a+π/4)
=sin(a+π/4)cos(a+π/4)
=1/2sin(2a+π/2)
=1/2cos2a

Answer 4

cos(π/4-a)cos(π/4+a)
=(cos(π/4)*cosa + sin(π/4)*sina) * (cos(π/4)*cosa - sin(π/4)*sina
=√2/2 * (cosa+ sina) * √2/2 * (cosa- sina)
=1/2 * (cos²a - sin²a)
=(cos2a)/2

Answer 5

结果为 1/2cos2a
cos（α+β）=cosαcosβ-sinαsinβ
cos（α-β）=cosαcosβ+sinαsinβ