化简cos(π/4-a)cos(π/4+a)
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化简cos(π/4-a)cos(π/4+a)
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)
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)
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