已知0<β<π/4<α<3π/4,cos(π/4-α)=3/5,sina(3π/4+β)=5/13,求cos(α-β﹚帮
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已知0<β<π/4<α<3π/4,cos(π/4-α)=3/5,sina(3π/4+β)=5/13,求cos(α-β﹚帮帮忙啊!
因为0<β<π/4<α<3π/4
所以-π/2<π/4-α<0
3π/4<3π/4+β<π
因为cos(π/4-α)=3/5,sin(3π/4+β)=5/13
所以sin(π/4-α)=-√[1-(3/5)^2]=-4/5
cos(3π/4+β)=-√[1-(5/13)^2]=-12/13
所以cos(α-β)=-cos[π-(α-β)]=-cos[(π/4-α)+(3π/4+β)]=-cos(π/4-α)cos(3π/4+β)+sin(π/4-α)sin(3π/4+β)=-(3/5)*(-12/13)+(-4/5)*(5/13)=16/65
如果不懂,请Hi我,祝学习愉快!
所以-π/2<π/4-α<0
3π/4<3π/4+β<π
因为cos(π/4-α)=3/5,sin(3π/4+β)=5/13
所以sin(π/4-α)=-√[1-(3/5)^2]=-4/5
cos(3π/4+β)=-√[1-(5/13)^2]=-12/13
所以cos(α-β)=-cos[π-(α-β)]=-cos[(π/4-α)+(3π/4+β)]=-cos(π/4-α)cos(3π/4+β)+sin(π/4-α)sin(3π/4+β)=-(3/5)*(-12/13)+(-4/5)*(5/13)=16/65
如果不懂,请Hi我,祝学习愉快!
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