If 0

Q: Wait, why did we assume $x = y$ in the final step? Is that always valid?

That is a fantastic question! In this specific problem, we are dealing with a constrained equation. When we simplified the expression to $\cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x}{2}\right) \sin\left(\frac{y}{2}\right) = \frac{1}{8}$, we are looking for values of $x$ and $y$ within $(0, \pi)$ that satisfy this. Because the maximum value of the product of these three functions is $\frac{1}{8}$ (which occurs at $x=y=60^\circ$), any deviation from this symmetry would make the product smaller than $\frac{1}{8}$. Thus, $x=y=60^\circ$ is the unique solution.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! However, this is a pure mathematical identity problem, not a physics problem. There is no 'energy' to conserve here, only the 'conservation' of trigonometric truth! We must rely on the rigid laws of trigonometry to navigate this.

Q: Why is the range $0 < x, y < \pi$ so important?

Great catch! The range $0 < x, y < \pi$ ensures that $\sin\left(\frac{x}{2}\right)$ and $\sin\left(\frac{y}{2}\right)$ are always positive. If $x$ or $y$ were outside this range, the signs could flip, potentially leading to different solutions or making the product negative. Always check your domain constraints—they are the boundaries of your mathematical playground!

Question

If $0 < x, y < \pi$ and $\cos x + \cos y - \cos(x+y) = \frac{3}{2}$, then $\sin x + \cos y$ is equal to:

Options:

(A) $\frac{1+\sqrt{3}}{2}$

(B) $\frac{1-\sqrt{3}}{2}$

(C) $\frac{\sqrt{3}}{2}$

(D) $\frac{1}{2}$

Sigma Percentile · Accepted Answer

### The Beauty of Trigonometric Symmetry

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are uncovering a hidden geometric truth.

We are presented with the equation $\cos x + \cos y - \cos(x+y) = \frac{3}{2}$, with the strict constraint that $0 < x, y < \pi$. At first glance, this looks like a tangled mess of variables. But in the world of JEE Advanced, complexity is often just a mask for elegance.

Let us peel back that mask together.

### Phase 1: The Art of Compression

Our first instinct should be to simplify. We have $\cos x + \cos y$ sitting right there.

Whenever you see a sum of cosines, your mind should immediately jump to the sum-to-product identity:
$$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}ight) \cos\left(\frac{C-D}{2}ight)$$

By applying this, we transform the sum into a product: $2 \cos\left(\frac{x+y}{2}ight) \cos\left(\frac{x-y}{2}ight)$. This is like folding a map—we have reduced the number of terms, making the landscape easier to navigate.

### Phase 2: The Universal Translator

Now, look at the third term: $-\cos(x+y)$. We have terms involving $\frac{x+y}{2}$ and $\frac{x-y}{2}$, but this third term is stuck in the language of $(x+y)$.

We need a translator. The half-angle identity $\cos 	heta = 2 \cos^2\left(\frac{	heta}{2}ight) - 1$ is our perfect bridge. By substituting $	heta = x+y$, we rewrite the term as $2 \cos^2\left(\frac{x+y}{2}ight) - 1$.

Now, every single term in our equation is speaking the same language of half-angles. This is the moment where the problem stops being a collection of random parts and starts becoming a cohesive system.

### Phase 3: The Elegant Cancellation

Let us assemble our pieces:
$$2 \cos\left(\frac{x+y}{2}ight) \cos\left(\frac{x-y}{2}ight) - \left[2 \cos^2\left(\frac{x+y}{2}ight) - 1ight] = \frac{3}{2}$$

When we distribute that negative sign and move the constant to the other side, we get:
$$2 \cos\left(\frac{x+y}{2}ight) \cos\left(\frac{x-y}{2}ight) - 2 \cos^2\left(\frac{x+y}{2}ight) = \frac{1}{2}$$

Notice the common factor? It is $2 \cos\left(\frac{x+y}{2}ight)$. Factoring it out leaves us with:
$$2 \cos\left(\frac{x+y}{2}ight) \left[\cos\left(\frac{x-y}{2}ight) - \cos\left(\frac{x+y}{2}ight)ight] = \frac{1}{2}$$

### Phase 4: The Final Revelation

Inside the bracket, we have another difference of cosines. Using the identity $\cos A - \cos B = 2 \sin\left(\frac{A+B}{2}ight) \sin\left(\frac{B-A}{2}ight)$, the bracket simplifies beautifully to $2 \sin\left(\frac{x}{2}ight) \sin\left(\frac{y}{2}ight)$.

Our equation is now:
$$4 \cos\left(\frac{x+y}{2}ight) \sin\left(\frac{x}{2}ight) \sin\left(\frac{y}{2}ight) = \frac{1}{2}$$

This simplifies to:
$$\cos\left(\frac{x+y}{2}ight) \sin\left(\frac{x}{2}ight) \sin\left(\frac{y}{2}ight) = \frac{1}{8}$$

This is the core of the problem. Given the constraints, this product reaches its maximum of $\frac{1}{8}$ only when $x = y = 60^\circ$.

Substituting these values into our target expression $\sin x + \cos y$, we get:
$$\sin(60^\circ) + \cos(60^\circ) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \mathbf{\frac{1+\sqrt{3}}{2}}$$

We have arrived at the finish line, and the path was nothing short of poetic.

Animated Solution for Mathematics - Trigonometry: If $0 < x, y < π$ and $cos x + cos y - cos (x + y) = \frac{3}{2}$ , then $sin x + cos y$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Equation

Apply Sum-to-Product Formula

Expand $cos (x + y)$ using Half-Angle

Substitute back into the Equation

Simplify the Constant Term

Factor out Common Terms

Apply Difference-to-Product Formula

Final Product Form

Deduce the Values of $x$ and $y$

Calculate the Target Expression

Final Result

The Sigma Insight: Trigonometric Ratios and Identities

The Beauty of Trigonometric Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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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 0<x,y<π and cosx+cosy−cos(x+y)=23​, then sinx+cosy is equal to: