Animated Solution for Mathematics - Trigonometry: In a triangle $ABC$, $2ac\sin \frac{1}{2}(A-B+C)=$

In any triangle $ABC$, we are tasked with evaluating the expression $2ac\sin \frac{1}{2}(A-B+C)$. While this expression appears complex, we can simplify it by leveraging the fundamental properties of triangles.

The first step is to utilize the angle sum property, which states that in any $△ABC$, the sum of the angles is $A+B+C=180^{\circ }$. This serves as our primary anchor for simplification.

We focus on the argument of the sine function: $\frac{1}{2}(A-B+C)$. By grouping the terms as $(A+C)-B$, we can substitute $A+C=180^{\circ }-B$ into the expression.

This transformation yields:

Distributing the $\frac{1}{2}$ across the terms, we obtain $90^{\circ }-B$. We are now evaluating $\sin (90^{\circ }-B)$.

Applying the complementary angle identity, $\sin (90^{\circ }-\theta )=\cos \theta$, the expression simplifies to $\cos B$. Consequently, our original expression $2ac\sin \frac{1}{2}(A-B+C)$ reduces to:

Whenever we encounter the product of two sides and the cosine of the included angle, we must invoke the Cosine Rule. The rule states:

Substituting this identity into our simplified expression, we get:

The term $2ac$ in the numerator and denominator cancels out perfectly. We are left with the final, elegant result: $a^2+c^2-b^2$