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定理 设n∈N,n>2,0<nx<π2,则sinnxsinx>n+3n.(1)证明:n=3时,应用sin3x=3sinx-4sin3x,0<x<π6,从而0<sin2x<14,即知(1)成立.设n=k时,(1)成立,sin(k+1)xsinx>k+1+3
Theorem Let n∈N,n>2,0n+3n. (1) Proof: When n=3, apply sin3x=3sinx-4sin3x, 0k+1+3