已知sin(3/4π+a)=5/13,cosa(1/4π-b)=3/5,且0<a<π/4<b<3/4π,求cos(a+b) 15
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cosa(1/4π-b)=3/5 是cos(1/4π-b)=3/5吧。
sin(a+b-π/2) = sin((3/4π+a)-(1/4π-b)) = sin(3/4π+a)*cos(1/4π-b)-cos(3/4π+a)*sin(1/4π-b)
sin(3/4π+a) = 5/13
cos(1/4π-b) = 3/5
cos(3/4π+a) = ±12/13
sin(1/4π-b) = ±4/5
因为0<a<π/4<b<3/4π，cos(3/4π+a) = -12/13，sin(1/4π-b) = 4/5
所以 cos(a+b) = -sin(a+b-π/2) = -(15/65 +48/65) = -63/65
sin(a+b-π/2) = sin((3/4π+a)-(1/4π-b)) = sin(3/4π+a)*cos(1/4π-b)-cos(3/4π+a)*sin(1/4π-b)
sin(3/4π+a) = 5/13
cos(1/4π-b) = 3/5
cos(3/4π+a) = ±12/13
sin(1/4π-b) = ±4/5
因为0<a<π/4<b<3/4π，cos(3/4π+a) = -12/13，sin(1/4π-b) = 4/5
所以 cos(a+b) = -sin(a+b-π/2) = -(15/65 +48/65) = -63/65
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