一道高一数学选择题
若对任意x属于R,cos(x+α)+cos(x+β)+cos(x+γ)=0,其中0<α<β<γ<2π,则γ-α=()A、π/3B、2π/3C、πD、4π/3写下过程,谢谢...
若对任意x属于R,cos(x+α)+cos(x+β)+cos(x+γ)=0,其中0<α<β<γ<2π,则γ-α=( )
A、π/3 B、2π/3 C、π D、4π/3
写下过程,谢谢!! 展开
A、π/3 B、2π/3 C、π D、4π/3
写下过程,谢谢!! 展开
2个回答
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领x=-y
cos(-y+α)+cos(-y+β)+cos(-y+γ)=0
cos(y-α)+cos(y-β)+1=0
领x=-α
cos(-α+α)+cos(-α+β)+cos(-α+γ)=0
=> 1+cos(-α+β)+cos(y-α)=0
=>cos(y-α)+cos(β-α)+1=0
领x=-β
cos(-β+α)+cos(-β+β)+cos(-β+γ)=0
=>cos(β-α)+1+cos(γ-β)=0
cos(y-α)+cos(y-β)+1=0 a
cos(y-α)+cos(β-α)+1=0 b
cos(β-α)+1+cos(γ-β)=0 c
a+b-c
2cos(y-α)=-1
cos(y-α)=-1/2
γ-α=2π/3
cos(-y+α)+cos(-y+β)+cos(-y+γ)=0
cos(y-α)+cos(y-β)+1=0
领x=-α
cos(-α+α)+cos(-α+β)+cos(-α+γ)=0
=> 1+cos(-α+β)+cos(y-α)=0
=>cos(y-α)+cos(β-α)+1=0
领x=-β
cos(-β+α)+cos(-β+β)+cos(-β+γ)=0
=>cos(β-α)+1+cos(γ-β)=0
cos(y-α)+cos(y-β)+1=0 a
cos(y-α)+cos(β-α)+1=0 b
cos(β-α)+1+cos(γ-β)=0 c
a+b-c
2cos(y-α)=-1
cos(y-α)=-1/2
γ-α=2π/3
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