State the Sine Rule forĀ  a triagle. Prove, in the usual notation for a triangle ABC. `(a^2-b^2)/c^2 = (sin(A-B))/(sin(A+B))`

1 Answer | Add Yours

aruv's profile pic

Posted on

In a triangle

`sin(A)/a=sin(B)/b=sin(C)/c`

This is known as Sine rule.

LHS=`(a^2-b^2)/c^`

`=(sin^2(A)-sin^2(B))/(sin^2(C))`

`=((sin(A)-sin(B))(sin(A)+sin(B)))/(sin^2(C))`

`=((2sin((A+B)/2)cos((A-B)/2))(2cos((A+B)/2)sin((A-B)/2)))/(sin^2(C))`

`=(sin(A+B)sin(A-B))/(sin^2(C))`

`A+B+C=180^0`

`A+B=180^o-C`

`sin(A+B)=sin(180^o-C)=sin(C)`

`therefore`

`LHS=(sin(C)sin(A-B))/(sin^2(C))`

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

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

`=RHS`

Hence proved.

We’ve answered 330,346 questions. We can answer yours, too.

Ask a question