For a triangle ABC it is given that cos A + cos B + cos C = 3/2. Prove that the triangle is equilate

Q: Why can't I just plug in $A=B=C=60^\circ$ and show it works? Isn't that a proof?

That is a verification, not a proof! In JEE Advanced, you must show that the equilateral case is the *only* solution. By using inequalities, we prove that the sum can never exceed $3/2$, and it hits $3/2$ only when the triangle is equilateral. We are proving uniqueness, not just existence.

Q: Why did we use $\cos C = 1 - 2 \sin^2 \frac{C}{2}$? Why not $\cos C = 2 \cos^2 \frac{C}{2} - 1$?

Great observation! We chose the sine version because our previous step generated a $\sin \frac{C}{2}$ term. We want consistency. If we used the cosine version, we would have mixed $\sin$ and $\cos$ half-angles, making the algebra a nightmare. Always aim for a single trigonometric function to simplify your life!

Q: How do we know $\sin \frac{C}{2}$ is positive? Could it be zero?

In a triangle, angles $A, B, C$ are strictly between $0$ and $\pi$. Therefore, $C/2$ is between $0$ and $\pi/2$. In this range, the sine function is always positive. We are safe to divide by it without worrying about division by zero!

Question

For a triangle $ABC$ it is given that $\cos A + \cos B + \cos C = \frac{3}{2}$. Prove that the triangle is equilateral.

Sigma Percentile · Accepted Answer

### The Mystery of the Three-Halves

Imagine you are standing in a vast, geometric landscape. You are given a triangle, $\Delta ABC$, and a single, cryptic clue: the sum of the cosines of its angles is exactly $\frac{3}{2}$.

That is, $\cos A + \cos B + \cos C = \frac{3}{2}$.

Your mission is to prove that this triangle is perfectly equilateral. This number is not arbitrary; it is the absolute maximum value that this sum can ever reach in a triangle.

### Phase 1: The Trigonometric Bridge

We start with our given condition: $\cos A + \cos B + \cos C = \frac{3}{2}$. We have three variables, $A$, $B$, and $C$, linked by the constraint $A + B + C = \pi$.

To make progress, we focus on the first two terms: $\cos A + \cos B$. We invoke the sum-to-product identity: $\cos X + \cos Y = 2 \cos \frac{X+Y}{2} \cos \frac{X-Y}{2}$.

Applying this to our expression, we obtain:

$$2 \cos \left( \frac{A+B}{2} ight) \cos \left( \frac{A-B}{2} ight) + \cos C = \frac{3}{2}$$

### Phase 2: The Geometric Constraint

Since $A + B + C = \pi$, we know that $A + B = \pi - C$. Dividing by two, we get $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.

Using the allied angle formula, $\cos(\frac{\pi}{2} - 	heta) = \sin 	heta$, we rewrite $\cos \frac{A+B}{2}$ as $\sin \frac{C}{2}$. Our equation now simplifies to:

$$2 \sin \left( \frac{C}{2} ight) \cos \left( \frac{A-B}{2} ight) + \cos C = \frac{3}{2}$$

### Phase 3: The Atomic Compute

To achieve total harmony, we convert $\cos C$ into a half-angle using the identity $\cos C = 1 - 2 \sin^2 \frac{C}{2}$. Substituting this into our equation, we get:

$$2 \sin \left( \frac{C}{2} ight) \cos \left( \frac{A-B}{2} ight) + 1 - 2 \sin^2 \left( \frac{C}{2} ight) = \frac{3}{2}$$

Isolating the term containing $A$ and $B$, we find:

$$2 \sin \left( \frac{C}{2} ight) \cos \left( \frac{A-B}{2} ight) = \frac{1}{2} + 2 \sin^2 \left( \frac{C}{2} ight)$$

Dividing by $2 \sin \frac{C}{2}$, we arrive at the pivotal expression:

$$\cos \left( \frac{A-B}{2} ight) = \frac{1 + 4 \sin^2 \left( \frac{C}{2} ight)}{4 \sin \left( \frac{C}{2} ight)}$$

### Phase 4: The Inequality Insight

We rewrite the right-hand side by completing the square. This allows us to manipulate the expression into the following form:

$$\cos \left( \frac{A-B}{2} ight) = 1 + \frac{(1 - 2 \sin \frac{C}{2})^2}{4 \sin \frac{C}{2}}$$

On the left, we have $\cos \frac{A-B}{2}$, which can never exceed $1$. On the right, we have $1$ plus a squared term divided by a positive value, meaning the right-hand side is always $\ge 1$.

The only way for a value $\le 1$ to equal a value $\ge 1$ is if both sides are exactly $1$. This forces $\cos \frac{A-B}{2} = 1$ (implying $A = B$) and $1 - 2 \sin \frac{C}{2} = 0$ (implying $\sin \frac{C}{2} = \frac{1}{2}$).

### Conclusion: The Equilateral Beauty

If $\sin \frac{C}{2} = \frac{1}{2}$, then $\frac{C}{2} = 30^\circ$, which means $C = 60^\circ$. Since we already established $A = B$, and the sum of angles is $180^\circ$, it follows that **$A = B = C = 60^\circ$**.

We have proven it! The triangle is **equilateral**.

Animated Solution for Mathematics - Trigonometry: For a triangle $A B C$ it is given that $cos A + cos B + cos C = \frac{3}{2}$ . Prove that the triangle is equilateral.

Visualized Solution

Given $cos A + cos B + cos C = \frac{3}{2}$

Sum-to-Product Identity

Raw Setup (Substitution)

Using $A + B + C = π$

Atomic Compute (Execution)

Double Angle Substitution

Substituting $cos C$

Isolating $cos \frac{A - B}{2}$

The Expression for Cosine

Completing the Square

Analyzing the Bounds

Solving for Equality

Conclusion: Equilateral Triangle

The Sigma Insight: Trigonometric Ratios and Identities

The Mystery of the Three-Halves

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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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; } For a triangle ABC it is given that cosA+cosB+cosC=23​. Prove that the triangle is equilateral.