观察题目观察等式sin20+sin40+sin20sin40=3/4
sin28+sin32+sin28sin32=3/4
(1)写出一个与以上两个等式规律相同的一个等式，并证明你的结论
(2)你能发现更一般的结论？并证明你的结论
解：因为20°+40°=28°+32°=60°,而cos60°=1/2,cos30°=√3/2
归纳到一般有：(1)若α+β=60°,则sin²α+sin²β+2sinαsinβ=3/4
(2)若α+β=γ,则sin²α+sin²β+2sinαsinβsinγ=sin²γ。
证明：(1)左边=(1-cos2α)/2+(1-cos2β)/2-[cos(α+β)-cos(α-β)]/2
`````````````=1-cos(α+β)cos(α-β)-cos(α+β)/2+cos(α-β)/2
`````````````=1-cos60°c...全部
解：因为20°+40°=28°+32°=60°,而cos60°=1/2,cos30°=√3/2
归纳到一般有：(1)若α+β=60°,则sin²α+sin²β+2sinαsinβ=3/4
(2)若α+β=γ,则sin²α+sin²β+2sinαsinβsinγ=sin²γ。
证明：(1)左边=(1-cos2α)/2+(1-cos2β)/2-[cos(α+β)-cos(α-β)]/2
`````````````=1-cos(α+β)cos(α-β)-cos(α+β)/2+cos(α-β)/2
`````````````=1-cos60°cos(α-β)-cos60°/2+cos(α-β)/2
`````````````=1-cos60°/2
`````````````=3/4=右边
(2)左边=(1-cos2α)/2+(1-cos2β)/2-[cos(α+β)-cos(α-β)]cosγ
`````````````=1-cos(α+β)cos(α-β)-cos(α+β)cosγ+cos(α-β)cosγ
`````````````=1-cosγcos(α-β)-cos²γ+cos(α-β)cosγ
`````````````=1-cos²γ
`````````````=sin²γ=右边
。
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