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

Concept:
To simplify expressions involving half–angles in a triangle, use the identities: \[ \sin^2\theta + \cos^2\theta = 1 \] \[ \cos^2\theta - \sin^2\theta = \cos 2\theta \] Also, the Cosine Rule for a triangle is: \[ a^2 = b^2 + c^2 - 2bc\cos A \]
Step 1: {Rewrite the given expression.} \[ (b-c)^2\cos^2\frac{A}{2} + (b+c)^2\sin^2\frac{A}{2} \] Expand the squares: \[ (b^2 + c^2 - 2bc)\cos^2\frac{A}{2} + (b^2 + c^2 + 2bc)\sin^2\frac{A}{2} \]
Step 2: {Group similar terms.} \[ (b^2+c^2)\left(\cos^2\frac{A}{2}+\sin^2\frac{A}{2}\right) -2bc\cos^2\frac{A}{2} +2bc\sin^2\frac{A}{2} \]
Step 3: {Apply trigonometric identities.} Since: \[ \cos^2\theta + \sin^2\theta = 1 \] the expression becomes: \[ (b^2+c^2) -2bc\left(\cos^2\frac{A}{2}-\sin^2\frac{A}{2}\right) \] Using: \[ \cos^2\theta - \sin^2\theta = \cos 2\theta \] \[ \cos^2\frac{A}{2}-\sin^2\frac{A}{2} = \cos A \] Thus the expression becomes: \[ b^2+c^2-2bc\cos A \]
Step 4: {Use the cosine rule.} \[ a^2 = b^2 + c^2 - 2bc\cos A \] Therefore, \[ (b-c)^2\cos^2\frac{A}{2} + (b+c)^2\sin^2\frac{A}{2} = a^2 \]
Step 5: {Conclusion.} Hence, the simplified value is: \[ a^2 \]