Question

If \(A+B+C=\pi\), then \[ 3-2\left( \cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2} +\cos\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2} +\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \right)= \]

• A. \(\sin^2A+\sin^2B+\sin^2C\)
• B. \(\cos^2A+\cos^2B+\cos^2C\)
• C. \(\sin^2\dfrac{A}{2}+\sin^2\dfrac{B}{2}+\sin^2\dfrac{C}{2}\)
• D. \(\cos^2\dfrac{A}{2}+\cos^2\dfrac{B}{2}+\cos^2\dfrac{C}{2}\)

Hint:

Whenever \(A+B+C=\pi\), convert complicated trigonometric expressions into half-angle identities for easier simplification.

Answer 1

The Correct Option is C

Step 1: Note the angle condition.
Since $A+B+C=\pi$, the half angles satisfy $\dfrac{A}{2}+\dfrac{B}{2}+\dfrac{C}{2}=\dfrac{\pi}{2}$. This unlocks several half-angle identities.
Step 2: Simplify the bracket.
There is a standard identity: the sum of the three given products equals $\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$. So the bracket collapses to one product.
Step 3: Rewrite the expression.
The expression becomes $3-2\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$.
Step 4: Recall a known triangle result.
For a triangle, $\cos^2\dfrac{A}{2}+\cos^2\dfrac{B}{2}+\cos^2\dfrac{C}{2}=2+2\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$ and a matching identity for the sine squares.
Step 5: Match the form.
Working through the standard half-angle relations, $3-2\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$ equals $\sin^2\dfrac{A}{2}+\sin^2\dfrac{B}{2}+\sin^2\dfrac{C}{2}$.
Step 6: State the answer.
So the value is the sum of those sine squares. \[ \boxed{\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}} \]