已知sinc+cosc=2sina,sinc*cosc=sin^b,求证:4cos^2 2a=cos^2 2b

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已知sinc+cosc=2sina,sinc*cosc=sin^b,求证:4cos^2 2a=cos^2 2b
已知sinc+cosc=2sina,sinc*cosc=sin^b,求证:4cos^2 2a=cos^2 2b

已知sinc+cosc=2sina,sinc*cosc=sin^b,求证:4cos^2 2a=cos^2 2b
由sinc+cosc=2sina平方可得1+2sinc*cosc=4sin^2 a
因sinc*cosc=sin^2 b 所以1+2sin^2 b=4sin^2 a
2-4sin^2 a=1-2sin^2 b
2cos2a=cos2b
平方后得4cos^2 2a=cos^2 2b