Question

若sinаcosβ=1/2则cosаsinβ的取值范围(要过程!)

Answer 1

sin(A+B) = sinA*cosB + cosAsinB

cosA*sinB = sin(A+B) - sinA*cosB = sin(A+B) - 1/2

因为 1≥sin(A+B)≥ -1

所以

1/2 ≥ cosAsinB ≥ -3/2

sin(A-B) = sinA*cosB - cosAsinB

cosA*sinB = sinA*cosB - sin(A-B) = 1/2 - sin(A-B)

3/2 ≥ cosAsinB ≥ -1/2

以上两结论取交集

1/2 ≥ cosAsinB ≥ -1/2

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