Question

If in a $\triangle ABC$, with usual notations, $a^2$, $b^2$, $c^2$ are in A.P., then $\frac{\sin 3B}{\sin B} = $

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

Hint:

When an identity problem involves an A.P. condition, pick easy values to find the answer quickly! Let $a^2=1$, $b^2=25$, and $c^2=49$ (so $a=1, b=5, c=7$). These form a right-angled or valid triangle structure where you can compute the values directly to verify matching exponent powers inside the denominator!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given that the squared side lengths $a^2$, $b^2$, and $c^2$ of a triangle are in an Arithmetic Progression (A.P.). We need to compute the corresponding value of the trigonometric sine ratio $\frac{\sin 3B}{\sin B}$.

Step 2: Key Formula or Approach:
Since $a^2$, $b^2$, and $c^2$ form an A.P., their terms satisfy the common difference balance: $$2b^2 = a^2 + c^2 \implies b^2 = \frac{a^2 + c^2}{2}$$ Recall the triple-angle identity for sine: $\sin 3B = 3\sin B - 4\sin^3 B$. Dividing this expression by $\sin B$ yields: $$\frac{\sin 3B}{\sin B} = 3 - 4\sin^2 B = 3 - 4(1 - \cos^2 B) = 4\cos^2 B - 1$$ We can substitute the Law of Cosines configuration for $\cos B$: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.

Step 3: Detailed Explanation:
Substitute the A.P. condition $b^2 = \frac{a^2 + c^2}{2}$ directly into the cosine expression: $$\cos B = \frac{a^2 + c^2 - \left(\frac{a^2 + c^2}{2}\right)}{2ac} = \frac{\frac{a^2 + c^2}{2}}{2ac} = \frac{a^2 + c^2}{4ac}$$ Now substitute this simplified cosine expression into our derived triple-angle layout: $$\frac{\sin 3B}{\sin B} = 4\cos^2 B - 1 = 4\left(\frac{a^2 + c^2}{4ac}\right)^2 - 1$$ $$= 4 \cdot \frac{(a^2 + c^2)^2}{16a^2c^2} - 1 = \frac{(a^2 + c^2)^2}{4a^2c^2} - 1$$ Find a common denominator to merge the terms: $$= \frac{(a^2 + c^2)^2 - 4a^2c^2}{4a^2c^2}$$ Expand the numerator using algebraic identity squares: $$= \frac{(a^4 + 2a^2c^2 + c^4) - 4a^2c^2}{4a^2c^2} = \frac{a^4 - 2a^2c^2 + c^4}{4a^2c^2}$$ Notice that the numerator matches the perfect square trinomial expansion of $(a^2 - c^2)^2$: $$= \frac{(a^2 - c^2)^2}{(2ac)^2} = \left(\frac{a^2 - c^2}{2ac}\right)^2$$ This matches option (B).

Step 4: Final Answer:
The ratio evaluates to $\left(\frac{a^2 - c^2}{2ac}\right)^2$, which corresponds to option (B).