Question

Обчисліть: a) sin 90° + sin 0°; B) tg 45°- tg 0°; I) cos 60° - sin 30°; 6) cos 90° – cos 0°; 1) tg 60° + cos 60°; A) sin 45° + cos 45º.

Answer 1

Ответ:

Объяснение:

a) sin 90° + sin 0°=1+0=1;

B) tg 45°- tg 0°=1-0=1;

I) cos 60° - sin 30°=1\2 -1\2=0;

6) cos 90° – cos 0°=0-1=-1;

1) tg 60° + cos 60°=√3 +1\2≈1,73+0,5≈2,23;

A) sin 45° + cos 45°=√2\2 + √2\2=  (2√2)\2=√2

бери значения из таблицы и подставляй-вот и вся хитрость.

Answer 2

a) sin 90° + sin 0°: sin 90° is equal to 1, and sin 0° is equal to 0. Therefore, the calculation is: 1 + 0 = 1.

b) tg 45° - tg 0°: tg 45° is equal to 1, and tg 0° is equal to 0. Therefore, the calculation is: 1 - 0 = 1.

I) cos 60° - sin 30°: cos 60° is equal to 0.5, and sin 30° is equal to 0.5. Therefore, the calculation is: 0.5 - 0.5 = 0.

6. cos 90° – cos 0°: cos 90° is equal to 0, and cos 0° is equal to 1. Therefore, the calculation is: 0 - 1 = -1.

7. tg 60° + cos 60°: tg 60° is equal to √3, and cos 60° is equal to 0.5. Therefore, the calculation is: √3 + 0.5 ≈ 2.366.

A) sin 45° + cos 45°: sin 45° is equal to √2/2, and cos 45° is equal to √2/2. Therefore, the calculation is: √2/2 + √2/2 = √2 ≈ 1.414.