Question

cossquare 48 - sinsquare12​

Answer 1

Answer:

question might be wrong

question : cos²48 - sin²42

=cos²48 - cos²(90-42)......................[sinθ = cos(90-θ)]

=cos²48 - cos²48

=0

:-)

Answer 2

Answer:

Step-by-step explanation:

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 cos 36° = 1/4(√5 +1)

= 1/2 × 1/4 (√5 + 1)

= (√5 + 1)/8