三角函数 证明试证 2sin^2 3θ-2sin^2 θ=cos2θ-cos6θ以θ=π/10代入上式,证明sin(3π
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三角函数 证明
试证 2sin^2 3θ-2sin^2 θ=cos2θ-cos6θ
以θ=π/10代入上式,证明sin(3π/10)-sin(π/10)=1/2
[用cosx=sin(π/2 - x)]
试证 2sin^2 3θ-2sin^2 θ=cos2θ-cos6θ
以θ=π/10代入上式,证明sin(3π/10)-sin(π/10)=1/2
[用cosx=sin(π/2 - x)]
因为
1- 2sin^2 θ=cos2 θ
所以2sin^2 3θ=1-cos6θ
-2sin^2 θ=-1+cos3θ
原式= 2sin^2 3θ-2sin^2 θ=cos2θ-cos6θ
把π/10带入
2sin²3π/10-2sin²2π/10=2(sin3π/10-sinπ/10)(sin3π/10+sinπ/10)=cos2π/10-cos6π/10
sin(3π/10)-sin(π/10)=(cos2π/10-cos6π/10)/2(sin3π/10+sinπ/10)=(sin3π/10-sin(-π/10))/2(sin3π/10+sinπ/10)=1/2
1- 2sin^2 θ=cos2 θ
所以2sin^2 3θ=1-cos6θ
-2sin^2 θ=-1+cos3θ
原式= 2sin^2 3θ-2sin^2 θ=cos2θ-cos6θ
把π/10带入
2sin²3π/10-2sin²2π/10=2(sin3π/10-sinπ/10)(sin3π/10+sinπ/10)=cos2π/10-cos6π/10
sin(3π/10)-sin(π/10)=(cos2π/10-cos6π/10)/2(sin3π/10+sinπ/10)=(sin3π/10-sin(-π/10))/2(sin3π/10+sinπ/10)=1/2
三角函数 证明试证 2sin^2 3θ-2sin^2 θ=cos2θ-cos6θ以θ=π/10代入上式,证明sin(3π
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