已知0°<β<45°<α<135°,cos(45°−α)=35,sin(135°+β)=513,求:
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已知0°<β<45°<α<135°,cos(45
解∵0°<β<45°<α<135°,
∴-90°<45°-α<0°,135°<135+β<180°
∵cos(45°−α)=
3
5,∴sin(45°−α)=−
4
5,
∵sin(135°+β)=
5
13,∴cos(135°+β)=−
12
13
∴sin(α+β)=-cos[(135°+β)-(45°-α)]
=-[cos(135°+β)cos(45°-α)+sin(135°+β)sin(45°-α)]
=−[(−
12
13)
3
5+
5
13(−
4
5)]=
56
65
cos(α-β)=-cos[(135°+β)+(45°-α)]
=[cos(135°+β)cos(45°-α)-sin(135°+β)sin(45°-α)]
=-[(−
12
13)
3
5−
5
13(−
4
5)]=
16
65
∴-90°<45°-α<0°,135°<135+β<180°
∵cos(45°−α)=
3
5,∴sin(45°−α)=−
4
5,
∵sin(135°+β)=
5
13,∴cos(135°+β)=−
12
13
∴sin(α+β)=-cos[(135°+β)-(45°-α)]
=-[cos(135°+β)cos(45°-α)+sin(135°+β)sin(45°-α)]
=−[(−
12
13)
3
5+
5
13(−
4
5)]=
56
65
cos(α-β)=-cos[(135°+β)+(45°-α)]
=[cos(135°+β)cos(45°-α)-sin(135°+β)sin(45°-α)]
=-[(−
12
13)
3
5−
5
13(−
4
5)]=
16
65
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