cos 48 - sin 12 =√5+1/8
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i think it should be cos^2 48 - sin^2 12...
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
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
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