Question

With usual notations, in \(△ABC\), \((b-c)^2cos^2\frac{A}{2}+(b+c)^2sin^2\frac{A}{2} =\)

• A. \(a^2\)
• B. \(b^2-c^2\)
• C. \(a^2+b^2+c^2\)
• D. \(0\)

Hint:

Expand both squares and use cos A = cos^2(A/2) - sin^2(A/2), then the cosine rule.

Answer 1

The Correct Option is A

Step 1: Test with a right triangle.
Take $A = 90^\circ$, $b = 3$, $c = 4$ (so $a = 5$). Then $\cos^2 45^\circ = \sin^2 45^\circ = 1/2$.

Step 2: Evaluate.
$(b - c)^2 \cdot \tfrac12 + (b + c)^2 \cdot \tfrac12 = \tfrac12(1 + 49) = 25 = a^2$.

Step 3: Confirm the identity.
Writing $\cos^2(A/2) = (1+\cos A)/2$ and $\sin^2(A/2) = (1-\cos A)/2$ and expanding gives $b^2 + c^2 - 2bc\cos A = a^2$ in general.

Final Answer:
Option (A). \[ \boxed{a^2} \]