Выражение: 9sin29° × sin225° × (sin^2 8° - cos^2 - вопрос №400892792922528281861458 от dcherbaev7771 30.08.2020 02:59

Question

Алгебра: Выражение: 9sin29° × sin225° × (sin^2 8° - cos^2 8°) + 18sin61° × cos45° × sin8°cos8°

dcherbaev7771 · Answer

9sin29° *sin225° *(sin²28°-cos²28°)+18° *sin61° *cos45° *sin28° *cos28°=4,5√2cos5°

1. sin29°=sin(90°-61°)=cos61°
2. sin225°=sin(180°+45°)=-sin45°=-√2/2
3. sin²28°-cos²28°=-(cos²28°-sin²28°)=-cos2*28°=-cos56°
4. 9cos61° *(-√2/2)*(-cos56°)=(9√2/2)*cos61° *cos56°

5. sin28° *cos28°=(1/2)*2*sin28° *cos28°=(1/2)sin(2*28°)=(1/2)sin56°

6. 18*sin61° *(√2/2)*(1/2)sin56°=(9√2/2)*sin61° *sin56°
7. (9√2/2)*cos61° *cos56°+(9√2/2)*sin61° *sin56°=(9√2/2)*(cos61° *cos56°+sin61° *sin56°)=(9√2/2)*cos(61°-56°)=(9√2/2)cos5°=4,5√2cos5°