已知sin(x-y)=3/5,sin(x+y)=-3/5,且x属于(兀/4,兀/2),y属于(-3兀/4,-兀/2),求
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已知sin(x-y)=3/5,sin(x+y)=-3/5,且x属于(兀/4,兀/2),y属于(-3兀/4,-兀/2),求sin2x的值.
解得:
x∈(π/4,π/2),y∈(-3π/4,-π/2)
∴x+y∈(-π/2,0),x-y∈(3π/4,5π/4)
又∵sin(x-y)=3/5,sin(x+y)=-3/5
∴cos²(x-y)=1-9/25=16/25
cos²(x+y)=1-9/25=16/25
∴cos(x-y)=-4/5,cos(x+y)=4/5
sin2x=sin[(x-y)+(x+y)]
=sin(x-y)cos(x+y)+sin(x+y)cos(x-y)
=3/5*4/5-3/5*(-4/5)
=24/25
∴sin2x的值为24/25.
x∈(π/4,π/2),y∈(-3π/4,-π/2)
∴x+y∈(-π/2,0),x-y∈(3π/4,5π/4)
又∵sin(x-y)=3/5,sin(x+y)=-3/5
∴cos²(x-y)=1-9/25=16/25
cos²(x+y)=1-9/25=16/25
∴cos(x-y)=-4/5,cos(x+y)=4/5
sin2x=sin[(x-y)+(x+y)]
=sin(x-y)cos(x+y)+sin(x+y)cos(x-y)
=3/5*4/5-3/5*(-4/5)
=24/25
∴sin2x的值为24/25.
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