Question

2(sin² 2 + cos 2) - 222 sin² 2+2 cos² 2 - log, 2² (sin² 2 + (052) 2 sin²2 +2c0s 2 - log, 2? ( sin'e ) - log 2: cos'2 sin²2 (2-log 2") + cos²2 ( 2-log 2") = sin² 2 (2log 2 - lag 22 ) + Cos² 2 ( 2 loy 2 - loy, 2² ) = sin^2 ( log, ?? - log, ??) + cos² 2 ( log₂ ?² = 2²) sin² 2 ₂ 22 22 + cos² 2 log₂ 22 22 sin'2 log, 1 + cos 2 log, sin 2.0+ 203² 2·0 = 0(sin² 2 + cos² 2) = ?​

Answer 1

2(1) - 2sin²(2) + 2cos²(2) - log₂(2²) (sin²(2) + (0.5²) sin²(2) + 2cos²(2) - log₂(2) (sin²(2)) - log₂(2) = 1 - 2sin²(2) + 2cos²(2) - 2(sin²(2)) - log₂(4) + 2(log₂(2) - log₂(2)²) - log₂(2) = 1 - 2sin²(2) + 2cos²(2) - 2sin²(2) - log₂(4) + 2 - 2 - log₂(2) = 1 - 2sin²(2) + 2cos²(2) - 2sin²(2) - log₂(4) = 1 - 4sin²(2) + 2cos²(2) - log₂(4) = 1 - 4sin²(2) + 2(1 - sin²(2)) - log₂(4) = 1 - 4sin²(2) + 2 - 2sin²(2) - log₂(4) = 3 - 6sin²(2) - log₂(4).