Let α and β be real numbers such that -π/4

Q: Why did we group the terms the way we did? Could we have just taken the LCM of all four at once?

Great question! While you technically could take the LCM of all four terms, it would lead to a massive, unmanageable expression in the numerator. By grouping them strategically, we force the appearance of the $\cos(\alpha - \beta)$ identity, which is the key to unlocking the entire problem. In JEE, look for symmetry—it is almost always a hint to simplify.

Q: I got stuck at the denominator $\sin 2\alpha \sin 2\beta$. How do I handle the product of two sine functions?

Don't panic! Whenever you see a product of sines or cosines, think of the product-to-sum identities. Specifically, $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$. By multiplying and dividing by 2, you can transform that product into a difference of cosines, which is much easier to evaluate using the values you already have.

Q: Wait, why do we use $\cos 2\theta = 2\cos^2\theta - 1$ for one part and $\cos 2\theta = 1 - 2\sin^2\theta$ for the other?

That is a brilliant observation! We choose the identity based on what we know. We are given $\cos(\alpha - \beta)$, so we use the identity that involves $\cos^2(\alpha - \beta)$. We are given $\sin(\alpha + \beta)$, so we use the identity that involves $\sin^2(\alpha + \beta)$. Always pick the tool that matches the data you have in your hand!

Question

Let $\alpha$ and $\beta$ be real numbers such that $-\frac{\pi}{4} < \beta < 0 < \alpha < \frac{\pi}{4}$. If $\sin(\alpha + \beta) = \frac{1}{3}$ and $\cos(\alpha - \beta) = \frac{2}{3}$, then the greatest integer less than or equal to $\left( \frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha} \right)^2$ is _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the trigonometric relations $\sin(\alpha + \beta) = \frac{1}{3}$ and $\cos(\alpha - \beta) = \frac{2}{3}$.

Our objective is to evaluate the expression:
$$E = \left( \frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha} ight)^2$$

Let the expression inside the square be $X$. To simplify $X$, we group the terms strategically:
$$X = \left( \frac{\sin \alpha}{\cos \beta} + \frac{\cos \alpha}{\sin \beta} ight) + \left( \frac{\cos \beta}{\sin \alpha} + \frac{\sin \beta}{\cos \alpha} ight)$$

### The Art of Strategic Grouping

By taking the least common multiple (LCM) for each pair, we observe:
$$X = \frac{\sin \alpha \sin \beta + \cos \alpha \cos \beta}{\cos \beta \sin \beta} + \frac{\cos \alpha \cos \beta + \sin \beta \sin \alpha}{\sin \alpha \cos \alpha}$$

Recognizing the identity $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$, we can factor out the common term:
$$X = \cos(\alpha - \beta) \left( \frac{1}{\sin \beta \cos \beta} + \frac{1}{\sin \alpha \cos \alpha} ight)$$

### Applying Double-Angle Identities

To simplify the denominators, we multiply the numerator and denominator of each fraction by $2$ to utilize the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$:
$$X = \cos(\alpha - \beta) \left( \frac{2}{\sin 2\beta} + \frac{2}{\sin 2\alpha} ight)$$

Combining these fractions yields:
$$X = 2 \cos(\alpha - \beta) \left( \frac{\sin 2\alpha + \sin 2\beta}{\sin 2\alpha \sin 2\beta} ight)$$

### Final Calculation

Using the sum-to-product identity $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$, the numerator becomes:
$$\sin 2\alpha + \sin 2\beta = 2 \sin(\alpha + \beta) \cos(\alpha - \beta)$$

Using the product-to-difference identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$, the denominator is:
$$\sin 2\alpha \sin 2\beta = \frac{1}{2} [\cos(2\alpha - 2\beta) - \cos(2\alpha + 2\beta)]$$

Substituting the known values $\sin(\alpha + \beta) = \frac{1}{3}$ and $\cos(\alpha - \beta) = \frac{2}{3}$ into the simplified expression, we find $X = -\frac{4}{3}$.

Squaring this result gives $E = \left( -\frac{4}{3} ight)^2 = \frac{16}{9} \approx 1.77$.

The greatest integer less than or equal to $1.77$ is **$1$**.

Enter Numerical Value:

Visualized Solution

Analyzing the Given Data

Grouping the Terms and Taking LCM

Applying Cosine Difference Formula

Factoring and Double Angles Setup

Sine Double Angle Formula

Sum to Product Formula

Product to Sum Formula

Double Angle Cosine Identities

Evaluating the Denominator

Evaluating the Numerator and $X$

Calculating Final Expression $E$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Art of Strategic Grouping

Applying Double-Angle Identities

Final Calculation

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