JEE Main 2026 22 January Morning Shift
Exam JEE Main
Year 2026
Date 22 January
Shift Morning Shift
Subject Mathematics
Chapter Trigonometric Ratios & Identities JEE Main 2026 · 22 January · Morning Shift · Mathematics · Trigonometric Ratios & Identities
If ^ 2 48^ - ^ 2 12^ ^ 2 24^ - ^ 2 6^ = α+β 5 2 , where α, β ∈ ℕ ,...
If
cos 2 4 8 ∘ − sin 2 1 2 ∘ sin 2 2 4 ∘ − sin 2 6 ∘ = α + β 5 2 \frac{\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}}{\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}}=\frac{\alpha+\beta \sqrt{5}}{2} s i n 2 2 4 ∘ − s i n 2 6 ∘ c o s 2 4 8 ∘ − s i n 2 1 2 ∘ = 2 α + β 5 , where
α , β ∈ N \alpha, \beta \in \mathbb{N} α , β ∈ N , then
α + β \alpha+\beta α + β is equal to
_ _ _ _ _ \_\_\_\_\_ _____
Step-by-step solution
Using
cos 2 A − sin 2 B = cos ( A + B ) cos ( A − B ) \cos^2A - \sin^2B = \cos(A+B)\cos(A-B) cos 2 A − sin 2 B = cos ( A + B ) cos ( A − B ) :
cos 2 48 ° − sin 2 12 ° = cos 60 ° cos 36 ° = 1 2 cos 36 ° \cos^2 48° - \sin^2 12° = \cos 60° \cos 36° = \frac{1}{2}\cos 36° cos 2 48° − sin 2 12° = cos 60° cos 36° = 2 1 cos 36° .
Using
sin 2 A − sin 2 B = sin ( A + B ) sin ( A − B ) \sin^2A - \sin^2B = \sin(A+B)\sin(A-B) sin 2 A − sin 2 B = sin ( A + B ) sin ( A − B ) :
sin 2 24 ° − sin 2 6 ° = sin 30 ° sin 18 ° = 1 2 sin 18 ° \sin^2 24° - \sin^2 6° = \sin 30° \sin 18° = \frac{1}{2}\sin 18° sin 2 24° − sin 2 6° = sin 30° sin 18° = 2 1 sin 18° .
Ratio
= cos 36 ° sin 18 ° = ( 1 + 5 ) / 4 ( 5 − 1 ) / 4 = ( 1 + 5 ) 2 4 = 6 + 2 5 4 = 3 + 5 2 = \frac{\cos 36°}{\sin 18°} = \frac{(1+\sqrt{5})/4}{(\sqrt{5}-1)/4} = \frac{(1+\sqrt{5})^2}{4} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2} = s i n 18° c o s 36° = ( 5 − 1 ) /4 ( 1 + 5 ) /4 = 4 ( 1 + 5 ) 2 = 4 6 + 2 5 = 2 3 + 5 .
α = 3 , β = 1 \alpha = 3,\ \beta = 1 α = 3 , β = 1 .
α + β = 4 \alpha + \beta = 4 α + β = 4 .
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