Question

Помогите пожалуйста с тригонометрией! Заранее благодарю)

Answer 1

1) √3/sin100° + 1/cos260° = √3/sin(90° + 10°) + 1/cos(270°- 10°) = √3/cos10° - 1/sin10° = (√3 · sin10° - 1 · cos10°)/(sin10° cos10°) = 2· ((√3/2) · sin10° - (1/2) · cos10°)/(sin10° cos10°) = 2· (cos30° · sin10° - sin30° · cos10°)/(sin10° cos10°) = 2· (sins(10° - 30°))/(sin10° cos10°) = -4· (sin20°)/(2·sin10° cos10°) = -4 · (sin20°)/sin20° = -4.

2) sin⁶α + cos⁶α + (3/4)sin²2α = (sin²α + cos²α)(sin⁴α - sin²αcos²α + cos⁴α) + (3/4)sin²2α = sin⁴α + 2sin²αcos²α + cos⁴α + (3/4)sin²2α - 3sin²αcos²α = (sin²α + cos²α)² + (3/4)sin²2α - 3sin²αcos²α = 1 + (3/4)sin²2α - (3/4)·4sin²αcos²α = 1 + (3/4)sin²2α - (3/4)sin²2α = 1