f(x)=2sin(x/3 一兀/6),x属于R 求f(0)
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f(x)=2sin(x/3 一兀/6),x属于R 求f(0)
设a,B属于[0,兀/2],f(3a+兀/2)=10/13,f(3B+2兀)=6/5 求sin(a+B)
设a,B属于[0,兀/2],f(3a+兀/2)=10/13,f(3B+2兀)=6/5 求sin(a+B)
f(0)=2sin[(0/3)-π/6]
=2sin(-π/6)
=-1
再问: 设a,B属于[0,兀/2],f(3a+兀/2)=10/13,f(3B+2兀)=6/5 求sin(a+B)
再答: f(3a+π/2)=2sin(a+π/6-π/6)=2sina=10/13 则: sina=5/13, 因:a∈[0,π/2],则:cosa=12/13 f(3b+2π)=2sin(b+2π/3-π/6)=2sin(b+π/2)=2cosb=6/5 则: cosb=3/5 因:b∈[0,π/2],则:sinb=4/5 sin(a+b)=sinacosb+cosasinb=63/65
=2sin(-π/6)
=-1
再问: 设a,B属于[0,兀/2],f(3a+兀/2)=10/13,f(3B+2兀)=6/5 求sin(a+B)
再答: f(3a+π/2)=2sin(a+π/6-π/6)=2sina=10/13 则: sina=5/13, 因:a∈[0,π/2],则:cosa=12/13 f(3b+2π)=2sin(b+2π/3-π/6)=2sin(b+π/2)=2cosb=6/5 则: cosb=3/5 因:b∈[0,π/2],则:sinb=4/5 sin(a+b)=sinacosb+cosasinb=63/65
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