(a)证明cos(x-y)-cos(x+y)=2sinxsiny (b)由此,证明 2sinθ(sinθ+sin3θ+sin5θ+sin7θ)=1-cos 8θ

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(a)证明cos(x-y)-cos(x+y)=2*sinx*siny (b)由此,证明 2sinθ(sinθ+sin3θ+sin5θ+sin7θ)=1-cos 8θ
(a)证明cos(x-y)-cos(x+y)=2*sinx*siny (b)由此,证明 2sinθ(sinθ+sin3θ+sin5θ+sin7θ)=1-cos 8θ

(a)证明cos(x-y)-cos(x+y)=2*sinx*siny (b)由此,证明 2sinθ(sinθ+sin3θ+sin5θ+sin7θ)=1-cos 8θ
(a)证明：
左边=cos(x-y)-cos(x+y)
=(cosxcosy+sinxsiny)-(cosxcosy-sinxsiny)
=2sinxsiny
=右边
证毕
(b)证明：
由(1)的结论知：
左边=2sinθ(sinθ+sin3θ+sin5θ+sin7θ)
=2sinθsinθ+2sinθsin3θ+2sinθsin5θ+2sinθsin7θ
=[cos(θ-θ)-cos(θ+θ)]+[cos(θ-3θ)-cos(θ+3θ)]+[cos(θ-5θ)-cos(θ+5θ)]+[cos(θ-7θ)-cos(θ+7θ)]
=cos0-cos2θ+cos2θ-cos4θ+cos4θ-cos6θ+cos6θ-cos8θ
=1-cos8θ
=右边
证毕

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