# In any triangle ABC prove that `(a-b)/b=(2*sin(C/2)*sin((A-B)/2))/sin B`

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In any triangle the law of sines gives `a/sin A = b/sin B = c/sin C = k`

We have to prove `(a-b)/b = (2*sin (C/2)*sin((A - B)/2))/sin B`

`(2*sin(C/2)*sin((A - B)/2))/sin B`

=> `(2*sin((180-(A+B))/2)*sin((A - B)/2))/(sin B)`

=> `(2*cos((A+B)/2)*sin((A - B)/2))/sin B`

=> `(sin((A+B)/2 + (A-B)/2) - sin((A+B)/2-(A - B)/2))/sin B`

=> `(sin A - sin B)/sin B`

Use the relation from the law of sines: `sin A = a/k` and `sin B = b/k`

=> `(a/k - b/k)/(b/k)`

=> `(a-b)/b`

**This proves that** `(a-b)/b = (2*sin(C/2)*sin((A-B)/2))/sin B`