Воспользуемся формулой sin(α ± β) = sinα∙cosβ ± cosα∙sinβ:
sin[arccos(3/5) + arcsin(8/17)] ⇔
⇔ sin[arccos(3/5)]∙cos[arcsin(8/17)] +
+ cos[arccos(3/5)]∙sin[arcsin(8/17)];
Воспользуемся формулами:
arcsin(x) = arccos√(1 - x²) при 0 ≤ x ≤ 1;
arccos(x) = arcsin√(1 - x²) при 0 ≤ x ≤ 1;
sin[arccos(3/5)]∙cos[arcsin(8/17)] +
+ cos[arccos(3/5)]∙sin[arcsin(8/17)] ⇔
⇔ sin{arcsin√[1 - (3/5)²]}∙cos{arccos√[1 - (8/17)²]} +
+ cos[arccos(3/5)]∙sin[arcsin(8/17)];
Вычислим отдельно:
√[1 - (3/5)²] = √[1 - 9/25] = √[16/25] = 4/5;
√[1 - (8/17)²] = √[1 - 64/289] = √[225/289] = 15/17;
Подставим в выражение:
sin{arcsin√[1 - (3/5)²]}∙cos{arccos√[1 - (8/17)²]} +
+ cos[arccos(3/5)]∙sin[arcsin(8/17)]) ⇔
⇔ sin[arcsin(4/5)]∙cos[arccos(15/17)] +
+ cos[arccos(3/5)]∙sin[arcsin(8/17)];
Теперь заменим:
sin[arcsin(α)] = α;
cos[arccos(α)] = α;
sin[arcsin(4/5)]∙cos[arccos(15/17)] +
+ cos[arccos(3/5)]∙sin[arcsin(8/17)] ⇔
⇔ (4/5)∙(15/17) + (3/5)∙(8/17) ⇔
⇔ 60/85 + 24/85 = 84/85.
Ответ: sin[arccos(3/5) + arcsin(8/17)] = 84/85.