已知sinA cosA=1/5,A∈(0,π),求sinA,cosA及sin^3A cos^3A 網友廻答:
- 1)sinA cosA=1/5.(1)
(sinA cosA)^2=1/25
1 2sinAcosa=1/25
sinAcosA=-12/25.(2)
(sinA-cosA)^2=1-2sinAcosA=1-2*(-12/25)=49/25
A∈(0,π),
sinA-cosA=±7/5
由(2)式知
cosA0
sinA-coaA=7/5.(3)
(1) (3)得
sinA=4/5
(1)-(2)得
cosA=-3/5
sin^3A cos^3A
=(sinA coaA)(sin^2A coa^2A sinAcosA)
=(4/5-3/5)(1-12/25)
=1/5*13/25
=13/125
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