Q.In $\triangle ABC$, $A+B+C=\pi$; show that $\sin^2\dfrac{A}{2}+\sin^2\dfrac{B}{2}-\sin^2\dfrac{C}{2}=1-2\cos\dfrac{A}{2}\cos\dfrac{B}{2}\sin\dfrac{C}{2}$

Pair $\sin2A+\sin2B$ using the sum-to-product formula (which brings in $\sin C$ via $A+B=\pi-C$), then combine with $\sin2C$ and use $\cos C=-\cos(A+B)$ to collapse everything to $4\sin A\sin B\sin C$.

Given: $A,B,C$ are angles of a triangle, so $A+B+C=\pi$. Show $\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$.

Step 1. Combine the first two terms using $\sin P+\sin Q = 2\sin\left(\dfrac{P+Q}{2}\right)\cos\left(\dfrac{P-Q}{2}\right)$:

$\sin2A+\sin2B = 2\sin(A+B)\cos(A-B)$

Step 2. Since $A+B=\pi-C$, $\sin(A+B)=\sin(\pi-C)=\sin C$:

$\sin2A+\sin2B = 2\sin C\cos(A-B)$

Step 3. Add $\sin2C = 2\sin C\cos C$:

$\sin2A+\sin2B+\sin2C = 2\sin C\cos(A-B) + 2\sin C\cos C = 2\sin C\left[\cos(A-B)+\cos C\right]$

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Combine $\sin A+\sin B$ via the sum-to-product formula (using $A+B=\pi-C$), write $\sin C=2\sin\frac{C}{2}\cos\frac{C}{2}$, then combine the two remaining cosine terms with another sum-to-product step.

Given $A+B+C=\pi$ (angles of a triangle).

Step 1 — combine $\sin A+\sin B$:

$\sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}$

Since $A+B=\pi-C$, we have $\dfrac{A+B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$, so $\sin\dfrac{A+B}{2}=\sin\!\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)=\cos\dfrac{C}{2}$. Thus:

$\sin A+\sin B=2\cos\dfrac{C}{2}\cos\dfrac{A-B}{2}$

Step 2 — write $\sin C$ using the double-angle formula:

$\sin C=2\sin\dfrac{C}{2}\cos\dfrac{C}{2}$

Step 3 — add and factor out $2\cos\dfrac{C}{2}$:

$\sin A+\sin B+\sin C=2\cos\dfrac{C}{2}\left[\cos\dfrac{A-B}{2}+\sin\dfrac{C}{2}\right]$

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