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: Understanding the Concept:
We need to combine the two terms using the half-angle identities and the cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\).

Step 2: Expand.
\[ (b-c)^2\cos^2\tfrac{A}{2} + (b+c)^2\sin^2\tfrac{A}{2} = (b^2 + c^2)\left(\cos^2\tfrac{A}{2} + \sin^2\tfrac{A}{2}\right) - 2bc\left(\cos^2\tfrac{A}{2} - \sin^2\tfrac{A}{2}\right) \]

Step 3: Simplify.
The first bracket is 1 and the second bracket is \(\cos A\).
\[ = b^2 + c^2 - 2bc\cos A = a^2 \]

Step 4: Check the options.
The result is \(a^2\), so (A). The other options do not equal \(b^2 + c^2 - 2bc\cos A\).

Final Answer:
The expression equals \(a^2\), option (A). \[ \boxed{a^2} \]