已知0<β<π/4,π/4<α<3π/4,cos(π/4-α)=3/5,sin(3π/4-β)=5/13,求cos(α+
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已知0<β<π/4,π/4<α<3π/4,cos(π/4-α)=3/5,sin(3π/4-β)=5/13,求cos(α+β)的值
答案是-56/65【要过程】
答案是-56/65【要过程】
π/4<α<3π/4,
-π/2<π/4-α<0
cos(π/4-α)=3/5
sin(π/4-α)=-4/5
0<β<π/4
π/2<3π/4-β<3π/4
sin(3π/4-β)=5/13
cos(3π/4-β)=-12/13
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]
=16/65
-π/2<π/4-α<0
cos(π/4-α)=3/5
sin(π/4-α)=-4/5
0<β<π/4
π/2<3π/4-β<3π/4
sin(3π/4-β)=5/13
cos(3π/4-β)=-12/13
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]
=16/65
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