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已知sinA/2=根号5/5,cos(A+B)=5/13,A属于(0,π),B属于(0,π/2) (1)求sin2A的值 (2)求sinB的值
人气:333 ℃ 时间:2020-04-08 11:16:13
解答
(1)sinA=2sinA/2*cosA/2,sinA/2=√5/5,A∈(0,π),cosA/2=2√5/5,sinA=2*√5/5*2√5/5=4/5
sin2A=2sinA*cosA,sinA=2*√5/5*2√5/5=4/5,cosA=3/5
=2*4/5*3/5=24/25.
(2)cos(A+B)=5/13,sinA=2*√5/5*2√5/5=4/5,sin2A=24/25,A属于(0,π),B属于(0,π/2)
sin(A+B)=12/13.
sinB=sin(A+B-A)=sin(A+B)*cosA-cos(A+B)*sinA=12/13*3/5-5/13*4/5=16/65
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