Question

With usual notations, in \(△ABC\), if \(2a^2 = b^2+c^2\), then \(\frac{cos3A}{cosA}+2 =\)

• A. \(\frac{b^2-c^2}{2bc}\)
• B. \((\frac{b^2-c^2}{2bc})^2\)
• C. \((\frac{c^2-b^2}{bc})^2\)
• D. \(\frac{c^2-b^2}{bc}\)

Hint:

Use cos 3A = 4cos^3 A - 3cos A to simplify the ratio, then use the cosine rule with the given relation.

Answer 1

The Correct Option is B

Step 1: Plan.
Reduce everything to $\cos A$, then express $\cos A$ in sides.

Step 2: Triple angle.
$\cos 3A = 4\cos^3A - 3\cos A$, so the expression is $4\cos^2 A - 1 = (2\cos A - 1)(2\cos A + 1)$.

Step 3: Use the sides.
$2\cos A = \dfrac{b^2 + c^2}{2bc}$. Then $2\cos A - 1 = \dfrac{(b - c)^2}{2bc}$ and $2\cos A + 1 = \dfrac{(b + c)^2}{2bc}$.

Step 4: Multiply.
\[ (2\cos A - 1)(2\cos A + 1) = \frac{(b-c)^2(b+c)^2}{4b^2c^2} = \left(\frac{b^2 - c^2}{2bc}\right)^2 \]

Final Answer:
Option (B). \[ \boxed{\left(\frac{b^2-c^2}{2bc}\right)^2} \]