Question

求证：cos^2x+cos^2(x+θ)-2cosxcosθcos(x+θ)=sin^2θ.

Answer 1

证明：cos²x+cos²(x+θ)-2cosxcosθcos(x+θ)
=cos²x+cos(x+θ)[cos(x+θ)-2cosxcosθ]
=cos²x+cos(x+θ)(-cosxcosθ-sinxsinθ)
=cos²x-(cosxcosθ-sinxsinθ)(cosxcosθ+sinxsinθ)
=cos²x-cos²xcos²θ+sin²xsin²θ
=cos²x(1-cos²θ)+sin²xsin²θ
=cos²xsin²θ+sin²xsin²θ
=sin²θ
所以等式得证。

Answer 2

cos^2x+cos^2(x+θ)-2cosxcosθcos(x+θ)
＝ cos^2x+cos(x+θ)[cos(x+θ)-2cosxcosθ]
=cos^2x+cos(x+θ)[cosxcosθ-sinxsinθ - -2cosxcosθ]
=cos^2x-cos(x+θ)[cosxcosθ+sinxsinθ]
=cos^2x-[cosxcosθ-sinxsinθ][cosxcosθ+sinxsinθ]
=cos^2x-cos^2xcos^2θ + sin^2xsin^2θ
=cos^2x-(1-sin^2θ)cos^2x + sin^2xsin^2θ
=sin^2θcos^2x+sin^2xsin^2θ
=sin^2θ

Answer 3

(cosx)^2+{cos(x+a)}^2-2cosxcosacos(x+a)=(cos2x+1)/2+(cos(2x+2a)+1)/2-2cosxcosacos(x+a)
=1+1/2{cos2x+cos(2x+2a)}-2cosxcosacos(x+a)
=1+1/2{2cos(2x+a)cosa}-2cosxcosacos(x+a)
=1+cosa{cos(2x+a)-2cosxcos(x+a)}
=1+cosa{cosxcos(x+a)-sinxsin(x+a)-2cosxcos(x+a)}
=1+cosa{-cos(x-x-a}
=1-(cosa)^2
=(sina)^2
参考:
cos(α+β)=cosα·cosβ-sinα·sinβ
cos(α-β)=cosα·cosβ+sinα·sinβ
sin(α±β)=sinα·cosβ±cosα·sinβ
tan(α+β)=(tanα+tanβ)/(1-tanα·tanβ)
tan(α-β)=(tanα-tanβ)/(1+tanα·tanβ)

·积化和差公式：
sinα·cosβ=(1/2)[sin(α+β)+sin(α-β)]
cosα·sinβ=(1/2)[sin(α+β)-sin(α-β)]
cosα·cosβ=(1/2)[cos(α+β)+cos(α-β)]
sinα·sinβ=-(1/2)[cos(α+β)-cos(α-β)]
·和差化积公式：
sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]
sinα-sinβ=2cos[(α+β)/2]sin[(α-β)/2]
cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]
cosα-cosβ=-2sin[(α+β)/2]sin[(α-β)/2]
如有步明白，可以追问