Question

In $\triangle ABC$, if $2a^2 = b^2 + c^2$, then the value of $\frac{\cos 3A}{\cos A} + 2$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Hint:

If a triangle problem gives a relation among sides and the answer options are constants, try substituting convenient values (such as an equilateral triangle when possible). This often simplifies the calculation quickly.

Answer 1

The Correct Option is A

Concept:
To simplify expressions involving multiple angles, use the triple–angle identity: \[ \cos 3A = 4\cos^3 A - 3\cos A \] Also, in a triangle the Cosine Rule gives: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Step 1: {Simplify the given expression.} \[ E = \frac{\cos 3A}{\cos A} + 2 \] Using the identity \( \cos 3A = 4\cos^3 A - 3\cos A \): \[ E = \frac{4\cos^3 A - 3\cos A}{\cos A} + 2 \] \[ E = 4\cos^2 A - 3 + 2 \] \[ E = 4\cos^2 A - 1 \]
Step 2: {Use the given condition.} Given: \[ 2a^2 = b^2 + c^2 \] This condition is satisfied by an equilateral triangle where: \[ a=b=c \] In an equilateral triangle: \[ A = 60^\circ \]
Step 3: {Substitute \(A = 60^\circ\).} \[ E = 4\cos^2 60^\circ - 1 \] \[ \cos 60^\circ = \frac{1}{2} \] \[ E = 4\left(\frac{1}{2}\right)^2 - 1 \] \[ E = 4\left(\frac{1}{4}\right) - 1 \] \[ E = 1 - 1 = 0 \]
Step 4: {Conclusion.} \[ \frac{\cos 3A}{\cos A} + 2 = 0 \]