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三道高一三角函数化简题1.cos(22°)*sin(38°)+s

2010-08-30 14:07:051***
1.cos(22°)*sin(38°)+sin(52°)*sin(22°) = 2.cos x sin(y-x)+cos(y-x) sin x= 3.将（2根号3）cosα-2sinα化为Asin(α+β)的形式为 要过程的，必要的步骤可写清所用公式，谢谢！三道高一三角函数化简题1.cos(22°)*sin(38°)+sin(52°)*sin(22°)=2.cosxsin(y-x)+cos(y-x)sinx=3.将?

最佳回答

  • 1. cos(22°)*sin(38°)+sin(52°)*sin(22°) =sin38cos22+cos38sin22 =sin(38+22) =sin60 =(√3)/2 2. cos x sin(y-x)+cos(y-x) sin x =sin(y-x)coxx+cos(y-x)sinx =sin(y-x+x) =siny 3.将（2根号3）cosα-2sinα =4[(√3)/2 cosa - (1/2)sina] =4[sin60°cosa-cos60°sina] =4sin(60°-a) =-4sin(α-60°)
    2010-08-30 14:28:22
  • cos(22°)*sin(38°)+sin(52°)*sin(22°) =cos(22°)*sin(38°)+cos(38°)*sin(22°) =cos(22°)*sin(38°)+sin(22°)*cos(38°) =sin(22°+38°) =根号3/2 cos x sin(y-x)+cos(y-x) sinx =cos x sin(y-x)+cos(x-y) sinx =sin(x+y-x) =siny （2根号3）cosα-2sinα =4[(根号3/2)cosα-1/2sinα] =4sin(60°-α)
    2010-08-30 14:26:37
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