已知α+β,α-β均为锐角且sin(α+β)=2√5/5,cos(α-β)=3√10/10,求sin2β的值要详解
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已知α+β,α-β均为锐角且sin(α+β)=2√5/5,cos(α-β)=3√10/10,求sin2β的值要详解
已知α+β,α-β均为锐角且sin(α+β)=2√5/5,cos(α-β)=3√10/10,求sin2β的值
要详解
已知α+β,α-β均为锐角且sin(α+β)=2√5/5,cos(α-β)=3√10/10,求sin2β的值要详解
sin2β=sin( (α+β)-(α-β) )=sin(α+β)cos(α-β)-cos(α+β)sin(α-β)
=2√5/5 * 3√10/10 - √5/5 * √10/10 = √2/2
sin2β=sin[(α+β)-(α-β)]=sin(α+β)cos(α-β)-cos(α+β)sin(α-β)=2√5/5x3√10/10-√5/5x√10/10
=30√2/50-5√2/50=√2/2
sin(α+β)=2√5/5,cos(α-β)=3√10/10
方便书写,设 m = 2√5/5, n = 3√10/10
因为α+β,α-β均为锐角
所以
α+β = acrsin(m)
α-β = arccos(n)
所以
β=(arcsin(m) - arccos(n))/2
2β = arcsin(m) - arccos(n)...
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sin(α+β)=2√5/5,cos(α-β)=3√10/10
方便书写,设 m = 2√5/5, n = 3√10/10
因为α+β,α-β均为锐角
所以
α+β = acrsin(m)
α-β = arccos(n)
所以
β=(arcsin(m) - arccos(n))/2
2β = arcsin(m) - arccos(n)
sin2β = sin(arcsinm - arccosn)
=sin(arcsinm)cos(arccosn) - cos(arcsinm)sin(arccosn)
=mn - cos(arccos(根号(1-m^2))sin(arcsin(根号(1-n^2))
=mn - 根号(1-m^2)*根号(1-n^2)
=3/5*根号(2) - 根号(5)/5 * 根号(10)/10
=6/10*根号(2) - 根号(2)/10
=1/2 * 根号(2)
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