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: Understanding the Concept:
The triple angle formula gives \(\cos 3A = 4\cos^3 A - 3\cos A\). Dividing by \(\cos A\) turns the ratio into an expression in \(\cos^2 A\) only.

Step 2: Simplify the ratio.
\[ \frac{\cos 3A}{\cos A} + 2 = 4\cos^2 A - 3 + 2 = 4\cos^2 A - 1 \]

Step 3: Find cos A.
Cosine rule: \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\). Since \(a^2 = \dfrac{b^2 + c^2}{2}\), the numerator is \(\dfrac{b^2 + c^2}{2}\) and
\[ \cos A = \frac{b^2 + c^2}{4bc} \]

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

Step 5: Check the options.
Option (B) matches. Option (C) has \(bc\) instead of \(2bc\) in the denominator. (A) and (D) are not squared.

Final Answer:
The value is \(\left(\dfrac{b^2 - c^2}{2bc}\right)^2\), option (B). \[ \boxed{\left(\frac{b^2-c^2}{2bc}\right)^2} \]