If (cos²48° − sin²12°)/(sin²24° − sin²6°) = (α + β√5)/2, where α, β ∈ N, then α + β is equal to

If (cos²48° − sin²12°)/(sin²24° − sin²6°) = (α + β√5)/2, where α, β ∈ N, then α + β is equal to | JEE Main Mathematics

NEETJEE

Math Trigonometry

Q Numerical Trigonometry

If $$ \frac{\cos^2 48^\circ – \sin^2 12^\circ}{\sin^2 24^\circ – \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, $$ where $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to ______

✅ Correct Answer

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Solution Steps

Convert cos²48°

$\cos 48^\circ = \sin 42^\circ$.

So numerator becomes: $\sin^2 42^\circ – \sin^2 12^\circ$.

Apply identity

$\sin^2 x – \sin^2 y = \sin(x+y)\sin(x-y)$.

Numerator = $\sin 54^\circ \sin 30^\circ = \frac{1}{2}\sin 54^\circ$.

Simplify denominator

$\sin^2 24^\circ – \sin^2 6^\circ = \sin 30^\circ \sin 18^\circ$.

= $\frac{1}{2}\sin 18^\circ$.

Take ratio

$\frac{\frac{1}{2}\sin54^\circ}{\frac{1}{2}\sin18^\circ} = \frac{\sin54^\circ}{\sin18^\circ}$.

Use exact values

$\sin54^\circ = \frac{\sqrt5 + 1}{4}$.

$\sin18^\circ = \frac{\sqrt5 – 1}{4}$.

Rationalize

$\frac{\sqrt5 + 1}{\sqrt5 – 1} = \frac{(\sqrt5+1)^2}{5-1} = \frac{6 + 2\sqrt5}{4} = \frac{3 + \sqrt5}{2}$.

Compare

$\alpha = 3$, $\beta = 1$.

$\alpha + \beta = \boxed{4}$.

1. Identity: Difference of Squares of Sine

The identity $\sin^2 x – \sin^2 y = \sin(x+y)\sin(x-y)$ is derived using product-to-sum formulas. Instead of expanding each squared term using double-angle identities, this formula directly converts the difference into a product form. This is extremely powerful in competitive exams like JEE Main because it reduces multiple steps into a single transformation. Whenever two squared sine terms are subtracted, students should immediately think of this identity. It avoids unnecessary expansion and keeps calculations clean and structured.

2. Converting Cosine into Sine

In trigonometric simplification problems, converting cosine into sine often makes expressions uniform. Using the complementary angle identity $\cos \theta = \sin(90^\circ – \theta)$ helps rewrite mixed expressions into only sine terms. In this question, $\cos 48^\circ$ was converted into $\sin 42^\circ$. This allowed the entire numerator to be written in terms of sine functions, making the difference identity applicable. Recognizing such angle relationships saves time and prevents algebraic complexity during exams.

3. Special Angles: 18°, 36°, 54°

Angles like $18^\circ$, $36^\circ$, and $54^\circ$ are closely related to the golden ratio. Their exact trigonometric values involve $\sqrt{5}$. For example, $\sin 18^\circ = \frac{\sqrt5 – 1}{4}$ and $\cos 36^\circ = \frac{\sqrt5 + 1}{4}$. These are standard results frequently used in JEE problems. Instead of calculating approximate decimal values, students must remember these radical forms. They help convert complicated ratios into simple algebraic expressions involving $\sqrt5$.

4. Rationalization Using Conjugates

When expressions contain radicals in the denominator, rationalization is required. This is done by multiplying numerator and denominator by the conjugate. For example, $\frac{\sqrt5 + 1}{\sqrt5 – 1}$ is simplified by multiplying with $\frac{\sqrt5 + 1}{\sqrt5 + 1}$. This eliminates the radical from the denominator and produces a clean algebraic form. Rationalization is an important algebraic technique used in both trigonometry and coordinate geometry.

5. Structure Recognition in JEE Trigonometry

Most JEE trigonometry questions are not about heavy computation but about recognizing patterns. If you see expressions like $\sin^2 A – \sin^2 B$, immediately think of product identities. If you see angles like 18°, 36°, or 54°, recall golden ratio values. If radicals appear in denominators, rationalize. JEE tests identity recognition and algebraic discipline more than raw calculation ability. Developing pattern recognition significantly improves speed and accuracy.

6. Strategy for Exact Value Problems

Exact value problems follow a structured approach: first simplify using identities, then convert all angles into related forms, substitute known exact values, and finally rationalize if needed. Never approximate angles unless explicitly required. Keeping expressions in radical form ensures precise answers that match required formats such as $\frac{\alpha + \beta\sqrt5}{2}$. This systematic approach is essential for scoring full marks in numerical-type trigonometry questions.

7. Importance of Algebraic Cleanliness

Trigonometric simplifications often become messy if steps are not organized properly. Writing intermediate expressions clearly, cancelling common factors like $\frac{1}{2}$ carefully, and simplifying before substitution prevents mistakes. Many students lose marks not because they do not know identities, but because they skip structured simplification. Maintaining algebraic clarity is crucial for competitive exams like JEE Main.

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FAQs

Which identity simplifies sin²x − sin²y?

The identity $\sin^2 x – \sin^2 y = \sin(x+y)\sin(x-y)$ is used. It converts a difference of squares into a product form. This greatly reduces algebraic complexity.

Why use sin54° and sin18°?

After applying identities, the expression reduces to $\frac{\sin54^\circ}{\sin18^\circ}$. These angles have exact radical values involving $\sqrt5$. That makes simplification clean and exact.

What is sin18° exactly?

$\sin18^\circ = \frac{\sqrt5 – 1}{4}$. This value comes from golden ratio relationships. It is frequently used in JEE trigonometry problems.

Why rationalize the denominator?

Rationalization removes radicals from the denominator. It helps match the required form $\frac{\alpha + \beta\sqrt5}{2}$. This step ensures a clean final expression.

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