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January 已知sinθ+cosθ=2sina,sinθ*cosθ=si
已知sinθ+cosθ=2sina,sinθ*cosθ=si
cos2β=1-2sin＾2β=1-2sinθ*cosθ
cos＾2 2β=1-4sinθ*cosθ+4(sinθ*cosθ)^2 (1)
由sinθ+cosθ=2sina得:1+2sinθ*cosθ=4sin^2a
可得sinθ*cosθ=(4sin^2a-1)/2,代入式(1),得
cos＾2 2β=1-2(4sin^2a-1)+(4sin^2a-1)^2
=1-8sin^2a+2+16sin^4a-8sin^2a+1=4(1-24in^2a)^2
=4cos＾2 2a,证毕.
最佳答案 - 由提问者2007-12-31 20:24:53选出
解答：由sinθ+cosθ=2sinα,( sinθ+cosθ) =4sin α,
即1+2sinθcosθ=4（1-cos2α）/2, 即2sinθcosθ=1-2cos2α,
由sinθcosθ=sin β,得2sinθcosθ=2sin β=1-cos2β,
∴1-2cos2α=1-cos2β, 即2cos2α=cos2β,两边平方得4cos 2α=cos 2β.
这题太简单，提示下，把第一式两边平方
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