Question

If in a \( \triangle ABC \), \( 2b^2 = a^2 + c^2 \), then
\[ \frac{\sin 3B}{\sin B} { is equal to:} \]

• A. \( \frac{c^2 - a^2}{2ca} \)
• B. \( \frac{c^2 - a^2}{ca} \)
• C. \( \frac{(c^2 - a^2)^2}{(ca)^2} \)
• D. \( \left( \frac{c^2 - a^2}{2ca} \right)^2 \)

Hint:

In problems involving trigonometric functions in triangles, use the law of cosines to relate sides and angles, and trigonometric identities to simplify expressions involving multiple angles.

Answer 1

The Correct Option is D

The expression for $\sin 3B$ in terms of $\sin B$ is derived as follows: \[ \frac{\sin 3B}{\sin B} = 3\sin B - 4\sin^3 B \] \[ = 3 - 4\sin^2 B \] \[ = 3 - 4(1 - \cos^2 B) \] \[ = -1 + 4\cos^2 B \] Substitute $\cos^2 B$ with $(a^2 + c^2 - b^2)^2 / (4a^2c^2)$: \[ = -1 + \frac{4(a^2 + c^2 - b^2)^2}{4a^2c^2} \] \[ = -1 + \frac{(a^2 + c^2 - b^2)^2}{a^2c^2} \] Simplify further: \[ = -1 + \left( \frac{a^2 + c^2}{2ac} \right)^2 \] \[ = -1 + \left( \frac{a^2 + c^2}{2ac} \right)^2 \] \[ = \left( \frac{c^2 - a^2}{2ac} \right)^2 \] Thus, the expression simplifies to: \[ \sin 3B = \sin B \left( \frac{c^2 - a^2}{2ac} \right)^2 \] Thus, the correct answer is Option D.