If in a ΔABC a cos²(C/2) + c cos²(A/2) = 3b/2, then the sides a, b and c
(a) satisfy a+b=c
(b) are i

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

This is a common confusion! In this problem, we are dealing with triangle properties, not energy conservation. We are using trigonometric identities to simplify the relationship between sides. There are no negative roots to reject because we are not solving a quadratic equation; we are simplifying a linear relationship between sides $a, b,$ and $c$.

Q: How do I know when to use the Projection Formula?

Great question! The Projection Formula, $b = a \cos C + c \cos A$, is your best friend whenever you see a linear combination of sides and cosines of the *other* angles. If you see $a \cos C + c \cos A$, it is almost always a signal to replace it with $b$.

Q: What if the question asked for $a, b, c$ in G.P. instead of A.P.?

If the result had been $b^2 = ac$, then they would be in G.P. The process remains the same: simplify the trigonometric expression until you get a relationship between the sides. The final algebraic form tells you the progression.

Question

If in a $\Delta ABC$ $a \cos^2\left(\frac{C}{2}\right) + c \cos^2\left(\frac{A}{2}\right) = \frac{3b}{2}$, then the sides $a, b$ and $c$

Options:

(A) satisfy $a+b=c$

(B) are in A.P

(C) are in G.P

(D) are in H.P.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the vertex $B$ of a triangle $\Delta ABC$. You are looking across at the sides $a$, $b$, and $c$.

The problem presents you with a seemingly daunting equation:

$$a \cos^2\left(\frac{C}{2}ight) + c \cos^2\left(\frac{A}{2}ight) = \frac{3b}{2}$$

In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask.

### The Half-Angle Obstacle

The first thing that catches our eye is the half-angle term. In triangle geometry, half-angles are rarely the final destination.

We recall the powerful identity:

$$\cos^2\left(\frac{	heta}{2}ight) = \frac{1 + \cos 	heta}{2}$$

By applying this to both $\cos^2\left(\frac{C}{2}ight)$ and $\cos^2\left(\frac{A}{2}ight)$, we transform our equation into:

$$a \left( \frac{1 + \cos C}{2} ight) + c \left( \frac{1 + \cos A}{2} ight) = \frac{3b}{2}$$

Suddenly, the half-angles are gone, replaced by the more familiar full angles.

### The Algebraic Expansion

Now, we must be precise. We multiply the entire equation by $2$ to clear the denominators, giving us:

$$a(1 + \cos C) + c(1 + \cos A) = 3b$$

Expanding this, we get:

$$a + a \cos C + c + c \cos A = 3b$$

Do not rush! Group the terms logically. We separate the constant sides from the trigonometric terms:

$$(a + c) + (a \cos C + c \cos A) = 3b$$

### The Geometric Soul: The Projection Formula

Here lies the heart of the problem. Look at the term $(a \cos C + c \cos A)$.

It is the classic **Projection Formula**. Geometrically, if you drop a perpendicular from vertex $B$ onto side $b$, you divide $b$ into two segments: $c \cos A$ and $a \cos C$.

Their sum is exactly $b$. Thus, we can replace that entire trigonometric bracket with a single variable: $b$. The equation collapses into:

$$(a + c) + b = 3b$$

### The Final Synthesis

The trigonometry has vanished, leaving us with a simple, beautiful algebraic relationship:

$$a + c = 2b$$

This is the definition of an Arithmetic Progression. If the sum of the first and third terms is twice the middle term, the sequence is in A.P.

Thus, the sides $a$, $b$, and $c$ are in **Arithmetic Progression**. We started with a complex trigonometric equation and ended with a fundamental property of sequences.

Animated Solution for Mathematics - Trigonometry: If in a $Δ A B C$ $a cos^{2} (\frac{C}{2}) + c cos^{2} (\frac{A}{2}) = \frac{3 b}{2}$ , then the sides $a, b$ and $c$

Select Answer:

Visualized Solution

Visualizing the Triangle $Δ A B C$

The Half-Angle Identity

Substituting the Identities

Clearing the Denominators

Expanding the Brackets

Grouping the Terms

The Projection Formula

Applying the Projection Formula

Final Simplification

Conclusion: Arithmetic Progression

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Half-Angle Obstacle

The Algebraic Expansion

The Geometric Soul

The Final Synthesis

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in a ΔABC acos2(2C​)+ccos2(2A​)=23b​, then the sides a,b and c