A) cos (arcsin 4 / 5 ) б ) sin(arctg 8/15) - вопрос №466528345815 от vprasckowya 12.05.2023 20:42

Question

Алгебра: A) cos (arcsin 4 / 5 ) б ) sin(arctg 8/15)

vprasckowya · Answer

U = arccos (- 4/5) v = arcsin 1/3 sin(u + v) = sin(arccos (- 4/5)).cos(arcsin 1/3) + cos(arccos (- 4/5)).sin(arcsin (1/3)) = = √[1 – (- 4/5)²] * √[1 – (1/3)²] + (- 4/5)*(1/3) = √[1 – 16/25] * √[1 – 1/9] - 4/15 = = √[(9/25)*(8/9)] – 4/15 = (2√2) / 5 - 4/15 = (6√2 - 4) / 15...