已知函数f(x)=cos²x+√3sinxcosx-1/2.已知cos(β-α)=4/5,cos(β+α)=-3/5,0
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![已知函数f(x)=cos²x+√3sinxcosx-1/2.已知cos(β-α)=4/5,cos(β+α)=-3/5,0](/uploads/image/z/3816820-28-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%26%23178%3Bx%2B%E2%88%9A3sinxcosx-1%2F2.%E5%B7%B2%E7%9F%A5cos%28%CE%B2-%CE%B1%EF%BC%89%3D4%2F5%2Ccos%28%CE%B2%2B%CE%B1%EF%BC%89%3D-3%2F5%2C0)
已知函数f(x)=cos²x+√3sinxcosx-1/2.已知cos(β-α)=4/5,cos(β+α)=-3/5,0
已知函数f(x)=cos²x+√3sinxcosx-1/2.已知cos(β-α)=4/5,cos(β+α)=-3/5,0
已知函数f(x)=cos²x+√3sinxcosx-1/2.已知cos(β-α)=4/5,cos(β+α)=-3/5,0
f(x)=cos²x+√3sinxcosx-1/2
=1/2cos(2x)+√3/2sin(2x)
=sin(2x+π/6)
cos(β-α)=4/5,cos(β+α)=-3/5,∵0