Question

【高一数学】三角函数的计算问题》》》

已知san(x-b)cosx-cos(b-x)sinx=3/5，b是第三象限角，求sin(b+5π/4)的值。

写出过程和答案，谢谢！

Answer 1

sin(x-b)cosx-cos(b-x)sinx
=sin(x-b)cosx-cos(x-b)sinx
=sin(x-b-x)
=sin-b=3/5
sinb=-3/5,b是第三象限，故而cosb=-4/5
sin(b+5π/4)
=sinbcos5π/4-sin5π/4cosb
=-sinbcosπ/4+sinπ/4cosb
=-0.5√2

Answer 2

sin(x-b)cosx-cos(b-x)sinx=sin(x-b)cosx-cos(x-b)sinx=sin[(x-b)-x]
=sin(-b)=-sinb=3/5;
sin(b+5π/4)=sinbcos(5pi/4)+cosbsin5pi/4)=-√2/2*（-3/5-4/5）
=7√2/10

Answer 3

你这是什么题啊？？？san是什么啊？？是sin还是tan??

Answer 4

s