已知cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,且β∈(0,π/4),α∈...

-α∈（-3π/4,-π/4）π/4-α∈（-π/2,0）sin(π/4 -α)=-4/5 sin(5π/4+β)=-12/13 sin(π+π/4+β)=-12/13 -sin(π/4+β)=-12/13 sin(π/4+β)=12/13 β∈（0，π/4）π/4+β∈（π/4，π/2）cos(π/4+β)=5/13 sin(β+α)=sin(π/4+β-π...

我的 已知cos(π/4 -α)=3/5,sin(5π/4+β)=-12/13,α∈(π/4,3π/4),β∈(0,π/4),求sin(α+β)的值 45 请在给出解答步骤后不要下线,我要提问我已经算到sin(π/4-α)=-4/5cos(π/4+β)=5/13了就是看下一步的答案看不懂了,为什么sin(α+β)=sin(α-π/4+B+π/4)?我...

由a∈(π/4,3π/4)得：π/4+a∈(π/2,π)故：cos(π/4-a)=sin(π/4+a)=4/5cos(π/4+a)=-3/5由b∈(0,π/4)得：π/4+b∈(π/4,π/2)又有：sin(5π/4+b)=-sin(π/4+b)=-12/13故：sin(π/4+b)=12/13cos(π/4+b)=√[1-(-1...

α∈(π/4,3π/4),β∈(0,π/4),( π/4-α ) ∈(-π/2,0 ） (5π/4+β)∈(5π/4,3π/2)所以有sin(π/4-α)=-√[1-(cos(π/4-α))^2]=-4/5 cox(5π/4+β)=-√[1-(sin(5π/4+β))^2]=-5/12sin(α+β）=-sin((5π/4+β)-(π/4-α))...

所以，sin(π/4-a)=-4/5 已知b∈(0,π/4)所以，5π/4+b∈(5π/4,3π/2)又，sin(5π/4+b)=-12/13 所以，cos(5π/4+b)=-5/13 而，sin[(5π/4+b)-(π/4-a)]=sin(π+a+b)=-sin(a+b)所以：sin(a+b)=-sin[(5π/4+b)-(π/4-a)]=-[sin(5π/4+b)*...

所以sin(π/4-a)<0 cos(π/4-a)=3/5,由(sin)^2+(cos)^2=1 所以sin(π/4-a)=-4/5 sin(5π/4+b)=sin(π+π/4+b)=-sin(π/4+b)=-12/13 sin(π/4+b)=12/13 0<b<π/4 所以π/4<π/4+b<π/2 所以cos(π/4+b)>0 所以cos(π/4+b)=5/13 sin(a+b)=...

∵β∈(0，π/4)，∴β+π/4∈(π/4,π/2)sin(4分之5π+β)=-12/13，即sin(β+π/4)=12/13，∴cos(β+π/4)=5/13 ∴sin(α+β)=sin[(α-π/4)+(β+π/4)]=sin(α-π/4)cos(β+π/4)+cos(α-π/4)sin(β+π/4)=4/5*5/13+3/5*12/13 =56/65 ...

这个已知条件有问题：由β∈（0,π/4），可以推知5π/4+β∈（5π/4,3π/2），所以，sin(5π/4+β)为负值(-5/13)，而不是正值（5/13）。计算方法：sin(α+β)=-sin(π+α+β)=-sin[(5π/4+β)-(π/4-α)]=-sin(5π/4+β)cos(π/4-α)+cos(5π/4+β)sin(π/4...

【参考答案】56/65 ∵α∈（π/4,3π/4），β∈（0,π/4）∴π/4 -a∈(-π/2, 0)，3π/4+β∈(3π/4, π)∴sin(π/4-a)=-√[1- (3/5)²]=-4/5 cos(3π/4+β)=-√[1-(5/13)²]=-12/13 ∴sin(a+β)=sin[(3π/4 +β)-(π/4- a)-(π/...

因为 cos(π/4-α)=sin(3π/4-α)=3/5 sin(3π/4+β)=5/13；所以 sin(α+β)=sin[(3π/4+β)-(3π/4-α)]=sin(3π/4+β)cos(3π/4-α)-cos(3π/4+β)sin(3π/4-α)=(5/13)*(4/5)-(12/13)*(3/5)=-16/65 ...