- 三角函数已知sinAcosB=1/2，求cosAsinB的取值范
- 已知sinAcB=1/2，求cosAsinB的取值范围
- sinAcosB+sinBcosA=sin(A+B)
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sinBcosA=sin(A+B)-1/2∈[-3/2,1/2] ........(1)
sinAcosB-sinBcosA=sin(A-B)
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sinBcosA=1/2-sin(A-B)=1/2+sin(B-A)∈[-1/2,3/2] ...(2)
(1),(2)取交集得，sinBcosA∈[-1/2,1/2]