The sides of a triangle inscribed in a given circle subtend angles α, β and γ at the centre. The min

Q: Wait, why is the sum of angles $2\pi$? Shouldn't it be $\pi$ for a triangle?

That is a classic trap! You are thinking of the interior angles of the triangle itself, which indeed sum to $\pi$. However, the problem explicitly states these are angles subtended at the *center* of the circle. The angles around a point in a plane always sum to a full rotation, which is $2\pi$ radians. Always read the geometry carefully!

Q: Why do we maximize the sum to minimize the mean? This feels counterintuitive.

It is perfectly natural to feel that way! Remember, we are dealing with a negative sign in front of the sum. If you have a value like $-10$ and $-2$, which is smaller? $-10$ is smaller. To make a number 'more negative' (smaller), you need to make the positive magnitude as large as possible. Maximizing the sum makes the final result as negative as possible.

Q: Is it always valid to assume $\alpha = \beta = \gamma$ for optimization?

In this specific case, yes. Because the function $f(\alpha, \beta, \gamma) = \sin \alpha + \sin \beta + \sin \gamma$ is symmetric with respect to all variables, and the constraint $\alpha + \beta + \gamma = 2\pi$ is also symmetric, the extremum must occur when the variables are equal. This is a powerful property of symmetric functions that saves you from complex calculus!

Question

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta$ and $\gamma$ at the centre. The minimum value of the arithmetic mean of $\cos (\alpha + \frac{\pi}{2}), \cos (\beta + \frac{\pi}{2})$ and $\cos (\gamma + \frac{\pi}{2})$ is equal to ..........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving a trigonometry problem; we are uncovering the hidden elegance of a triangle inscribed in a circle. Imagine standing at the center of a circle, watching the three sides of an inscribed triangle. Each side casts a shadow, an angle, at your feet. We call these angles $\alpha, \beta,$ and $\gamma$.

The first step in any great geometric journey is to define our boundaries. Since these angles surround the center point $O$, they must complete a full rotation. Thus, our fundamental constraint is:

$$\alpha + \beta + \gamma = 2\pi$$

Keep this equation close; it is the key that will unlock the entire problem.

### The Transformation: Simplifying the Complexity

The problem asks us to find the minimum value of the arithmetic mean of three terms: $\cos(\alpha + \frac{\pi}{2}), \cos(\beta + \frac{\pi}{2}),$ and $\cos(\gamma + \frac{\pi}{2})$. At first glance, this looks intimidating. But let us breathe.

We have a powerful tool in our trigonometric arsenal: the allied angle identity. We know that $\cos(	heta + \frac{\pi}{2}) = -\sin 	heta$. This identity is a gift; it shifts our perspective from cosine to sine and introduces a negative sign.

Applying this to our expression, the arithmetic mean becomes:

$$AM = \frac{1}{3} [-\sin \alpha - \sin \beta - \sin \gamma]$$

We can factor out that negative sign to get:

$$AM = -\frac{1}{3} [\sin \alpha + \sin \beta + \sin \gamma]$$

Now, the problem transforms. We are no longer looking for the minimum of a complex cosine expression; we are looking for the minimum of a negative sum of sines.

### The Optimization Trap: Thinking Like a Mathematician

Here is where many students stumble. We want to minimize the arithmetic mean. But look at the expression: it is negative.

To make a number as small as possible—meaning as negative as possible—we must make the positive part, the sum $S = \sin \alpha + \sin \beta + \sin \gamma$, as large as possible. This is the 'negative sign trap.' We are not minimizing the sum; we are maximizing it to push the mean further into the negative.

So, how do we maximize $\sin \alpha + \sin \beta + \sin \gamma$ subject to $\alpha + \beta + \gamma = 2\pi$? This is a beautiful application of symmetry. In a system where the variables are interchangeable, the maximum value occurs when the variables are equal. If we set $\alpha = \beta = \gamma$, then $3\alpha = 2\pi$, which means each angle must be $\frac{2\pi}{3}$.

### The Final Calculation: Bringing it Home

With $\alpha = \beta = \gamma = \frac{2\pi}{3}$, we calculate the maximum sum:

$$S_{max} = \sin\left(\frac{2\pi}{3}ight) + \sin\left(\frac{2\pi}{3}ight) + \sin\left(\frac{2\pi}{3}ight)$$

We know that $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$. Therefore:

$$S_{max} = 3 	imes \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$$

Now, we return to our arithmetic mean expression: $AM_{min} = -\frac{1}{3} 	imes S_{max}$. Substituting our value:

$$AM_{min} = -\frac{1}{3} 	imes \frac{3\sqrt{3}}{2}$$

The threes cancel out with elegant precision, leaving us with the final answer:

**$$-\frac{\sqrt{3}}{2}$$**

You see? When you break the problem down into its geometric soul, the complexity vanishes. You have mastered the symmetry, navigated the negative sign, and arrived at the truth.

Animated Solution for Mathematics - Trigonometry: The sides of a triangle inscribed in a given circle subtend angles $α, β$ and $γ$ at the centre. The minimum value of the arithmetic mean of $cos (α + \frac{π}{2}), cos (β + \frac{π}{2})$ and $cos (γ + \frac{π}{2})$ is equal to ..........

Visualized Solution

Geometric Setup

Angle Sum at Center

The Arithmetic Mean

Applying Trigonometric Identity

Simplified AM Expression

Minimizing the Mean

Maximizing $sin α + sin β + sin γ$

Equating the Angles

Evaluating $S_{ma x}$

Finding the Minimum AM

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Transformation

The Optimization Trap

The Final Calculation

Similar Questions

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.

The minimum value of the expression $sin α + sin β + sin γ$ , where $α, β, γ$ are real numbers satisfying $α + β + γ = π$ is

If $cos (β - γ) + cos (γ - α) + cos (α - β) = - 3/2$ , then

$A B C$ is a triangle such that $sin (2 A + B) = sin (C - A) = - sin (B + 2 C) = 1/2$ . If $A, B$ and $C$ are in arithmetic progression, determine the values of $A, B$ and $C$ .

In the figure, $θ_{1} + θ_{2} = \frac{π}{2}$ and $3 (B E) = 4 (A B)$ . If the area of $Δ C A B$ is $23 - 3 unit^{2}$ , when $\frac{θ _{2}}{θ _{1}}$ is the largest, then the perimeter (in unit) of $Δ C E D$ is equal to

Let in a right angled triangle, the smallest angle be $θ$ . If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then $sin θ$ is equal to:

Let $∣ cos θ cos (6 0^{\circ} - θ) cos (6 0^{\circ} + θ) ∣ \leq \frac{1}{8}, θ \in [0, 2 π]$ . Then, the sum of all $θ \in [0, 2 π]$ , where $cos 3 θ$ attains its maximum value, is :

Given $α + β - γ = π$ , prove that $sin^{2} α + sin^{2} β - sin^{2} γ = 2 sin α sin β cos γ$ .

If $I_{n}$ is the area of $n$ sided regular polygon inscribed in a circle of unit radius and $O_{n}$ be the area of the polygon circumscribing the given circle, prove that $I_{n} = \frac{O _{n}}{2} (1 + 1 - (\frac{2 I _{n}}{n})^{2})$ .