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已知sin(α+3π/4)=5/13,cos(π/4-β)=4/5,且-π/4<α<π/4,π/4
已知sin(α+3π/4)=5/13,cos(π/4-β)=4/5,且-π/4<α<π/4,π/4<β<3π/4,求cos(α-β)的值...
已知sin(α+3π/4)=5/13,cos(π/4-β)=4/5,且-π/4<α<π/4,π/4<β<3π/4, 求cos(α-β)的值
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解:
注意到:
(α+3π/4)+(π/4-β)=π+(α-β)
因此,羡乱欲求cos(α-β),利用cos(α-β)=-cos[π+(α-β)]=-cos[(π/4-β)+(α+3π/4)],可求解
∵兄握档-π/4<α<π/4
∴π/2<α+3π/皮竖4<π
cos(α+3π/4)<0
∴cos(α+3π/4)=-12/13
又∵π/4<β<3π/4
即:-π/2<π/4-β<0
sin(π/4-β)<0
∴sin(π/4-β)=-3/5
cos(α-β)=-cos[π+(α-β)]=-cos[(π/4-β)+(α+3π/4)]
=sin(π/4-β)sin(α+3π/4)-cos(π/4-β)cos(α+3π/4)
=(-3/5)×(5/13)-(4/5)×(-12/13)
=33/65
注意到:
(α+3π/4)+(π/4-β)=π+(α-β)
因此,羡乱欲求cos(α-β),利用cos(α-β)=-cos[π+(α-β)]=-cos[(π/4-β)+(α+3π/4)],可求解
∵兄握档-π/4<α<π/4
∴π/2<α+3π/皮竖4<π
cos(α+3π/4)<0
∴cos(α+3π/4)=-12/13
又∵π/4<β<3π/4
即:-π/2<π/4-β<0
sin(π/4-β)<0
∴sin(π/4-β)=-3/5
cos(α-β)=-cos[π+(α-β)]=-cos[(π/4-β)+(α+3π/4)]
=sin(π/4-β)sin(α+3π/4)-cos(π/4-β)cos(α+3π/4)
=(-3/5)×(5/13)-(4/5)×(-12/13)
=33/65
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sin(a+3π/4)=5/13,其中,π/2<a+3π/4<π,则:cos(a+3π/4)=-12/13
cos(π/4-b)=4/5,其中键芹哪,-π/2<π/4-b<0,则:sin(π/4-b)=-首者3/5
则:
cos(a-b)
=-cos[π+(a-b)]
=-cos[(a+3π/4)+(π/4-b)]
=-[cos(a+稿码3π/4)cos(π/4-b)-sin(a+3π/4)sin(π/4-b)]
=-[(-12/13)×(4/5)-(5/13)×(-3/5)]
=33/65
cos(π/4-b)=4/5,其中键芹哪,-π/2<π/4-b<0,则:sin(π/4-b)=-首者3/5
则:
cos(a-b)
=-cos[π+(a-b)]
=-cos[(a+3π/4)+(π/4-b)]
=-[cos(a+稿码3π/4)cos(π/4-b)-sin(a+3π/4)sin(π/4-b)]
=-[(-12/13)×(4/5)-(5/13)×(-3/5)]
=33/65
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因为-π/4<α<π/4,π/4<β<3/4π,所以π/2<α+3/4π<π,-π/2<π/4-β<0
cos(α+3/4π)=12/13, sin(π/氏培4-β)=3/5
cos(α-β)薯核局=cos((α+3/4π)+(π/4-β)),然后利用公式和上面的值便可数让得出答案
cos(α+3/4π)=12/13, sin(π/氏培4-β)=3/5
cos(α-β)薯核局=cos((α+3/4π)+(π/4-β)),然后利用公式和上面的值便可数让得出答案
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