Question

In $\triangle ABC$, $(b-c)^2 \cos^2 \frac{A}{2} + (b+c)^2 \sin^2 \frac{A}{2} =$

• A. $a$
• B. $a^2$
• C. $b^2 + c^2$
• D. $2a^2$

Hint:

In triangle identities involving half–angles, first expand algebraic terms and then apply identities like \(\sin^2\theta+\cos^2\theta=1\) and \(\cos^2\theta-\sin^2\theta=\cos2\theta\). Most such expressions finally reduce to the cosine rule \(a^2=b^2+c^2-2bc\cos A\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to simplify a trigonometric expression involving the sides and half-angles of a triangle.
Step 3: Detailed Explanation:
Expand the given expression:
\[ (b^2 + c^2 - 2bc) \cos^2 \frac{A}{2} + (b^2 + c^2 + 2bc) \sin^2 \frac{A}{2} \]
Rearrange terms:
\[ (b^2 + c^2) \left( \cos^2 \frac{A}{2} + \sin^2 \frac{A}{2} \right) - 2bc \cos^2 \frac{A}{2} + 2bc \sin^2 \frac{A}{2} \]
Since $\cos^2 \theta + \sin^2 \theta = 1$:
\[ (b^2 + c^2)(1) - 2bc \left( \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2} \right) \]
Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$, we have $\cos A = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2}$:
\[ b^2 + c^2 - 2bc \cos A \]
By the Cosine Rule, $a^2 = b^2 + c^2 - 2bc \cos A$.
So, the expression equals $a^2$.
Step 4: Final Answer:
The simplified value is $a^2$.