Question

If the angles $A$, $B$, and $C$ of a triangle are in an arithmetic progression and if $a$, $b$, and $c$ denote the lengths of the sides opposite to $A$, $B$, and $C$ respectively, then the value of the expression $\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$ is

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. 1
• D. $\sqrt{3}$

Hint:

Trigonometry Tip: Whenever angles of a triangle are in an A.P., immediately lock in the middle angle $B = 60^\circ$. It is a universal rule that will save you significant time.

Answer 1

The Correct Option is D

Concept: Trigonometry - Properties of Triangles (Sine Rule) and Arithmetic Progressions.

Step 1: Determine angle B from the Arithmetic Progression property. Given that angles A, B, and C are in an Arithmetic Progression (A.P.), the middle term is the average of the other two: $2B = A + C$.
In any triangle, the sum of interior angles is $180^\circ$: $A + B + C = 180^\circ$. Substitute $A + C$ with $2B$: $(2B) + B = 180^\circ \implies 3B = 180^\circ$. Therefore, angle $B = 60^\circ$.

Step 2: Apply the Sine Rule to the given expression. The Sine Rule states that the ratio of a side length to the sine of its opposite angle is constant: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. This implies that $a$ is proportional to $\sin A$, and $c$ is proportional to $\sin C$. Thus, we can rewrite the ratios as $\frac{a}{c} = \frac{\sin A}{\sin C}$ and $\frac{c}{a} = \frac{\sin C}{\sin A}$.
Substitute these into our target expression: $\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A = \left(\frac{\sin A}{\sin C}\right)\sin 2C + \left(\frac{\sin C}{\sin A}\right)\sin 2A$.

Step 3: Expand the double angles using trigonometric identities. Use the double angle formula for sine, $\sin 2\theta = 2\sin\theta\cos\theta$, to expand the expression: $\left(\frac{\sin A}{\sin C}\right)(2\sin C\cos C) + \left(\frac{\sin C}{\sin A}\right)(2\sin A\cos A)$.

Step 4: Cancel terms and simplify the expression. In the first term, $\sin C$ cancels out from the numerator and denominator. In the second term, $\sin A$ cancels out. This leaves: $2\sin A\cos C + 2\sin C\cos A$. Factor out the 2: $2(\sin A\cos C + \cos A\sin C)$.

Step 5: Use the angle sum formula to evaluate. The expression inside the parentheses exactly matches the angle addition formula for sine: $\sin(X+Y) = \sin X\cos Y + \cos X\sin Y$.
So, $2(\sin A\cos C + \cos A\sin C) = 2\sin(A + C)$. From Step 1, we know $A + C = 2B$. Since $B = 60^\circ$, $A + C = 120^\circ$.
Substitute this angle: $2\sin(120^\circ) = 2\sin(180^\circ - 60^\circ) = 2\sin(60^\circ)$. Using standard trigonometric values, $\sin(60^\circ) = \frac{\sqrt{3}}{2}$. Evaluate: $2 \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}$. $$ \therefore \text{The value of the expression is } \sqrt{3}. $$