91Ó°ÊÓ

We know from the sine rule that in any triangle ABC, we have $$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} $$ Rearrange the equation to express the sides of the triangle in terms of the angles: $$ a = k\sin A,\quad b = k\sin B, \quad c = k\sin C $$ where k is some constant.

Now we will substitute the expressions of a, b, and c that we found in the previous step into the given expression: $$ k\sin A(\sin B-\sin C)+k\sin B(\sin C-\sin A)+k\sin C(\sin A-\sin B)=0 $$

Since k is a constant, we can factor it out of the entire expression: $$ k[\sin A(\sin B-\sin C)+\sin B(\sin C-\sin A)+\sin C(\sin A-\sin B)]=0 $$

We will now apply the sum-to-product formulas for the sine function to simplify our expression: The sum-to-product formula for sine is $$ \sin x - \sin y = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) $$ Using this, we can rewrite the expressions: $$ \sin A (\sin B - \sin C) = 2\cos\dfrac{A+B}{2}\sin\dfrac{B-C}{2} $$ Similar expressions can be written for the other two terms: $$ \sin B (\sin C - \sin A) = 2\cos\dfrac{B+C}{2}\sin\dfrac{C-A}{2} \\ \sin C (\sin A - \sin B) = 2\cos\dfrac{C+A}{2}\sin\dfrac{A-B}{2} $$ Substitute these expressions back into the expression found in step 3: $$ k[2\cos\dfrac{A+B}{2}\sin\dfrac{B-C}{2} + 2\cos\dfrac{B+C}{2}\sin\dfrac{C-A}{2}+ 2\cos\dfrac{C+A}{2}\sin\dfrac{A-B}{2}]=0 $$

Now, divide the entire expression by 2: $$ k[\cos\dfrac{A+B}{2}\sin\dfrac{B-C}{2}+\cos\dfrac{B+C}{2}\sin\dfrac{C-A}{2}+\cos\dfrac{C+A}{2}\sin\dfrac{A-B}{2}]=0 $$ We know that the sum of angles of a triangle is 180 degrees or \(\pi\) radians, so \(A+B+C=\pi\). We can substitute this relationship into the above expression: $$ k[\cos\dfrac{A+(180^\circ-A)}{2}\sin\dfrac{(180^\circ-A)-C}{2}+\cos\dfrac{(180^\circ-A)+A}{2}\sin\dfrac{(180^\circ-A)-A}{2}+\cos\dfrac{(180^\circ-A)-(180^\circ-A)}{2}\sin\dfrac{A-(180^\circ-A)}{2}]=0 $$ Simplify the expression by substituting the angles: $$ k[\cos90^\circ\sin\dfrac{180^\circ-C}{2}+\cos90^\circ\sin\dfrac{180^\circ-A}{2}+\cos0^\circ\sin\dfrac{A}{2}]=0 $$ We know \(\cos90^{\circ}=0\), \(\cos 0^{\circ}=1\), resulting in: $$k[0\cdot\sin\dfrac{180^\circ-C}{2}+0\cdot\sin\dfrac{180^\circ-A}{2}+1\cdot\sin\dfrac{A}{2}]=k\sin\frac{A}{2}=0$$ Since sine can never be equal to 0 for a triangle with non-zero angles, the constant k must be equal to 0. Therefore, the given expression is true for any triangle ABC.