Question

Cos ^2 48- sin ^2 12

Answer 1

cos²48° - sin²12°
we know ,
sin(90 - ∅) = cos∅
so,
sin12° = sin(90-78) = cos78°

cos²48° - cos²78°

[use, a² - b² = (a - b)(a + b) ]

= (cos48° + cos78°)(cos48°-cos78°)

[ use, formula ,
cosC +cosD = 2cos(C+ D)/2.cos(C-D)/2
cosC - cosD = 2sin(C +D)/2 .sin(D - C)/2 ]

= 2cos63°.cos15°.2sin63°.sin15°

[use, 2sinA.cosA = sin2A ]

=sin30.sin126°

= 1/2 sin(90+36°)

=1/2. cos36°

put cos36° = 1/4(√5 +1)

= 1/2 × 1/4 (√5 + 1)

= (√5 + 1)/8

Answer 2

Given cos^2 48 - sin^2 12

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

=cos(48+12)cos(48-12)

=cos60.cos36

=1/2.cos36

{Use : cos36=_/5 +1/4}

=1/2._/5+1/4

=_/5+1/8