JEE Main 2026 (22 January Shift 1)

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Animated Solution for Mathematics - Trigonometry: If $\frac{c o s ^{2} 4 8 ^{\circ} - s i n ^{2} 1 2 ^{\circ}}{s i n ^{2} 2 4 ^{\circ} - s i n ^{2} 6 ^{\circ}} = \frac{α + β 5}{2}$ , where $α, β \in N$ , then $α + β$ is equal to .........

Q: How do I memorize $\cos 36^\circ$ and $\sin 18^\circ$?

Don't just memorize them! Think of a triangle with angles $36^\circ, 72^\circ, 72^\circ$. By drawing an angle bisector, you create similar triangles that lead directly to the golden ratio, $\phi = \frac{1+\sqrt{5}}{2}$. These values are the backbone of many JEE problems, so keep them in your 'mental toolkit'.

Enter Numerical Value:

Visualized Solution

Identity for Numerator

Identify the numerator expression: $cos^{2} 4 8^{\circ} - sin^{2} 1 2^{\circ}$
Use the identity: $cos^{2} A - sin^{2} B = cos (A + B) cos (A - B)$

Simplify Numerator

Substitute $A = 4 8^{\circ}$ and $B = 1 2^{\circ}$
$cos^{2} 4 8^{\circ} - sin^{2} 1 2^{\circ} = cos (4 8^{\circ} + 1 2^{\circ}) cos (4 8^{\circ} - 1 2^{\circ})$
Result: $cos 6 0^{\circ} cos 3 6^{\circ}$

Identity for Denominator

Identify the denominator expression: $sin^{2} 2 4^{\circ} - sin^{2} 6^{\circ}$
Use the identity: $sin^{2} A - sin^{2} B = sin (A + B) sin (A - B)$

Simplify Denominator

Substitute $A = 2 4^{\circ}$ and $B = 6^{\circ}$
$sin^{2} 2 4^{\circ} - sin^{2} 6^{\circ} = sin (2 4^{\circ} + 6^{\circ}) sin (2 4^{\circ} - 6^{\circ})$
Result: $sin 3 0^{\circ} sin 1 8^{\circ}$

Form the Ratio

Form the ratio: $\frac{c o s 6 0 ^{\circ} c o s 3 6 ^{\circ}}{s i n 3 0 ^{\circ} s i n 1 8 ^{\circ}}$
Substitute $cos 6 0^{\circ} = \frac{1}{2}$ and $sin 3 0^{\circ} = \frac{1}{2}$

Simplify the Ratio

Cancel out the common factor of $\frac{1}{2}$
The ratio simplifies to: $\frac{c o s 3 6 ^{\circ}}{s i n 1 8 ^{\circ}}$

Recall Standard Values

Recall standard trigonometric values:
$cos 3 6^{\circ} = \frac{5 + 1}{4}$
$sin 1 8^{\circ} = \frac{5 - 1}{4}$

Substitute Values

Substitute the values into the ratio: $\frac{\frac{5 + 1}{4}}{\frac{5 - 1}{4}}$
Cancel the denominators ( $4$ ) to get: $\frac{5 + 1}{5 - 1}$

Rationalize Denominator

Multiply numerator and denominator by the conjugate $5 + 1$
$\frac{5 + 1}{5 - 1} \times \frac{5 + 1}{5 + 1}$

Simplify Expression

Numerator: $(5 + 1)^{2} = 5 + 1 + 25 = 6 + 25$
Denominator: $(5)^{2} - 1^{2} = 5 - 1 = 4$
Resulting fraction: $\frac{6 + 2 5}{4}$

Final Form Comparison

Divide by $2$ : $\frac{3 + 5}{2}$
Compare with the given form: $\frac{α + β 5}{2}$

Calculate $α + β$

From comparison, $α = 3$ and $β = 1$
Final calculation: $α + β = 3 + 1 = 4$

00:00 / 00:00

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Numerator

The expression provided is:

\frac{cos ^{2} 4 8 ^{\circ} - sin ^{2} 1 2 ^{\circ}}{sin ^{2} 2 4 ^{\circ} - sin ^{2} 6 ^{\circ}}

We begin by simplifying the numerator:

cos^{2} 4 8^{\circ} - sin^{2} 1 2^{\circ}

. Using the trigonometric identity

cos^{2} A - sin^{2} B = cos (A + B) cos (A - B)

, we substitute

A = 4 8^{\circ}

and

B = 1 2^{\circ}

This yields:

cos (4 8^{\circ} + 1 2^{\circ}) cos (4 8^{\circ} - 1 2^{\circ}) = cos 6 0^{\circ} cos 3 6^{\circ}

The Denominator's Turn

Next, we address the denominator:

sin^{2} 2 4^{\circ} - sin^{2} 6^{\circ}

. We apply the identity

sin^{2} A - sin^{2} B = sin (A + B) sin (A - B)

Substituting

A = 2 4^{\circ}

and

B = 6^{\circ}

, we obtain:

sin (2 4^{\circ} + 6^{\circ}) sin (2 4^{\circ} - 6^{\circ}) = sin 3 0^{\circ} sin 1 8^{\circ}

The Magic of Cancellation

Now, we combine the simplified numerator and denominator into the ratio:

\frac{cos 6 0 ^{\circ} cos 3 6 ^{\circ}}{sin 3 0 ^{\circ} sin 1 8 ^{\circ}}

Since

cos 6 0^{\circ} = \frac{1}{2}

and

sin 3 0^{\circ} = \frac{1}{2}

, these terms cancel out perfectly. We are left with the expression:

\frac{cos 3 6 ^{\circ}}{sin 1 8 ^{\circ}}

The Final Rationalization

We substitute the standard JEE-essential values:

cos 3 6^{\circ} = \frac{5 + 1}{4}

and

sin 1 8^{\circ} = \frac{5 - 1}{4}

. The expression becomes:

\frac{\frac{5 + 1}{4}}{\frac{5 - 1}{4}} = \frac{5 + 1}{5 - 1}

To rationalize, we multiply the numerator and denominator by the conjugate

(5 + 1)

\frac{( 5 + 1 ) ^{2}}{( 5 ) ^{2} - 1 ^{2}} = \frac{5 + 1 + 2 5}{4} = \frac{6 + 2 5}{4}

Simplifying by dividing by

2

, we arrive at:

\frac{3 + 5}{2}

Comparing this to the form

\frac{α + β 5}{2}

, we identify

α = 3

and

β = 1

. Therefore, the final result is:

α + β = 3 + 1 = 4

Common Traps & FAQs

Wait, why can't we just use

cos 2 θ

formulas?

You absolutely could, but it would lead to a labyrinth of half-angle formulas that might confuse you. The product identities

cos^{2} A - sin^{2} B = cos (A + B) cos (A - B)

are the 'pro' way to handle these specific forms. They turn squares into products, which is exactly what we need for cancellation.

How do I memorize

cos 3 6^{\circ}

and

sin 1 8^{\circ}

Don't just memorize them! Think of a triangle with angles

3 6^{\circ}, 7 2^{\circ}, 7 2^{\circ}

. By drawing an angle bisector, you create similar triangles that lead directly to the golden ratio,

ϕ = \frac{1 + 5}{2}

. These values are the backbone of many JEE problems, so keep them in your 'mental toolkit'.

Why do we rationalize the denominator?

In mathematics, especially in competitive exams like JEE, we prefer the 'standard form' where the denominator is a rational number. It makes comparing your result with the given form

\frac{α + β 5}{2}

straightforward. Without rationalizing, you'd be guessing the values of

α

and

β

Animated Solution for Mathematics - Trigonometry: If $\frac{c o s ^{2} 4 8 ^{\circ} - s i n ^{2} 1 2 ^{\circ}}{s i n ^{2} 2 4 ^{\circ} - s i n ^{2} 6 ^{\circ}} = \frac{α + β 5}{2}$ , where $α, β \in N$ , then $α + β$ is equal to .........

Enter Numerical Value:

Visualized Solution

Identity for Numerator

Simplify Numerator

Identity for Denominator

Simplify Denominator

Form the Ratio

Simplify the Ratio

Recall Standard Values

Substitute Values

Rationalize Denominator

Simplify Expression

Final Form Comparison

Calculate $α + β$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Numerator

The Denominator's Turn

The Magic of Cancellation

The Final Rationalization

Similar Questions

The value of $2 sin (1 2^{\circ}) - sin (7 2^{\circ})$ is :

If $sin^{4} α + 4 cos^{4} β + 2 = 42 sin α cos β$ ; $α, β \in [0, π]$ , then $cos (α + β) - cos (α - β)$ is equal to :

Let $cos (α + β) = - \frac{1}{10}$ and $sin (α - β) = \frac{3}{8}$ , where $0 < α < \frac{π}{3}$ and $0 < β < \frac{π}{4}$ . If $tan 2 α = \frac{3 ( 1 - r 5 )}{11 ( s + 5 )}, r, s \in N$ , then $r + s$ is equal to ......... .

If $\sum_{r = 1}^{13} {\frac{1}{s i n ( \frac{π}{4} + ( r - 1 ) \frac{π}{6} ) s i n ( \frac{π}{4} + \frac{r π}{6} )}} = a 3 + b,$ a, $b \in Z,$ then $a^{2} + b^{2}$ is equal to :

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

If the solution of the equation $lo g_{c o s x} cot x + 4 lo g_{s i n x} tan x = 1, x \in (0, \frac{π}{2})$ is $sin^{- 1} (\frac{α + β}{2})$ , where $α, β$ are integers, then $α + β$ is equal to:

If the value of $\frac{3 c o s 3 6 ^{\circ} + 5 s i n 1 8 ^{\circ}}{5 c o s 3 6 ^{\circ} - 3 s i n 1 8 ^{\circ}}$ is $\frac{a 5 - b}{c}$ , where $a, b, c$ are natural numbers and $g cd (a, c) = 1$ , then $a + b + c$ is equal to :

The expression $3 [sin^{4} (\frac{3 π}{2} - α) + sin^{4} (3 π + α)] - 2 [sin^{6} (\frac{π}{2} + α) + sin^{6} (5 π - α)]$ is equal to

Let $α$ and $β$ respectively be the maximum and the minimum values of the function $f (θ) = 4 (sin^{4} (\frac{7 π}{2} - θ) + sin^{4} (11 π + θ)) - 2 (sin^{6} (\frac{3 π}{2} - θ) + sin^{6} (9 π - θ)), θ \in R$ . Then $α + 2 β$ is equal to :

If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to:

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin224∘−sin26∘cos248∘−sin212∘​=2α+β5​​, where α,β∈N, then α+β is equal to .........

Enter Numerical Value:

Visualized Solution

Identity for Numerator

Simplify Numerator

Identity for Denominator

Simplify Denominator

Form the Ratio

Simplify the Ratio

Recall Standard Values

Substitute Values

Rationalize Denominator

Simplify Expression

Final Form Comparison

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+β

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Numerator

The Denominator's Turn

The Magic of Cancellation

The Final Rationalization

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the value of 5cos36∘−3sin18∘3cos36∘+5sin18∘​ is ca5​−b​, where a,b,c are natural numbers and gcd(a,c)=1, then a+b+c is equal to :

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Animated Solution for Mathematics - Trigonometry: If $\frac{c o s ^{2} 4 8 ^{\circ} - s i n ^{2} 1 2 ^{\circ}}{s i n ^{2} 2 4 ^{\circ} - s i n ^{2} 6 ^{\circ}} = \frac{α + β 5}{2}$ , where $α, β \in N$ , then $α + β$ is equal to .........

Calculate $α + β$

The value of $2 sin (1 2^{\circ}) - sin (7 2^{\circ})$ is :

If $sin^{4} α + 4 cos^{4} β + 2 = 42 sin α cos β$ ; $α, β \in [0, π]$ , then $cos (α + β) - cos (α - β)$ is equal to :

Let $cos (α + β) = - \frac{1}{10}$ and $sin (α - β) = \frac{3}{8}$ , where $0 < α < \frac{π}{3}$ and $0 < β < \frac{π}{4}$ . If $tan 2 α = \frac{3 ( 1 - r 5 )}{11 ( s + 5 )}, r, s \in N$ , then $r + s$ is equal to ......... .

If $\sum_{r = 1}^{13} {\frac{1}{s i n ( \frac{π}{4} + ( r - 1 ) \frac{π}{6} ) s i n ( \frac{π}{4} + \frac{r π}{6} )}} = a 3 + b,$ a, $b \in Z,$ then $a^{2} + b^{2}$ is equal to :

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

If the solution of the equation $lo g_{c o s x} cot x + 4 lo g_{s i n x} tan x = 1, x \in (0, \frac{π}{2})$ is $sin^{- 1} (\frac{α + β}{2})$ , where $α, β$ are integers, then $α + β$ is equal to:

If the value of $\frac{3 c o s 3 6 ^{\circ} + 5 s i n 1 8 ^{\circ}}{5 c o s 3 6 ^{\circ} - 3 s i n 1 8 ^{\circ}}$ is $\frac{a 5 - b}{c}$ , where $a, b, c$ are natural numbers and $g cd (a, c) = 1$ , then $a + b + c$ is equal to :

The expression $3 [sin^{4} (\frac{3 π}{2} - α) + sin^{4} (3 π + α)] - 2 [sin^{6} (\frac{π}{2} + α) + sin^{6} (5 π - α)]$ is equal to

If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to: