Chadwick Patel
03/03/2023 · Senior High School
86. Қимати ифодахоро ёбед: \( \sin 300^{\circ} ; \quad \cos 330^{\circ} ; \quad \sin 570^{\circ} \);
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Tutor-Verified Answer
Step-by-step Solution
Давайте найдем значения указанных тригонометрических функций.
1. \( \sin 300^{\circ} \):
Угол \( 300^{\circ} \) находится в четвертой четверти. Мы можем использовать соотношение:
\[
\sin(300^{\circ}) = -\sin(360^{\circ} - 300^{\circ}) = -\sin(60^{\circ}) = -\frac{\sqrt{3}}{2}
\]
2. \( \cos 330^{\circ} \):
Угол \( 330^{\circ} \) также находится в четвертой четверти. Используем аналогичное соотношение:
\[
\cos(330^{\circ}) = \cos(360^{\circ} - 330^{\circ}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}
\]
3. \( \sin 570^{\circ} \):
Угол \( 570^{\circ} \) можно привести к углу в пределах \( 0^{\circ} \) и \( 360^{\circ} \):
\[
570^{\circ} - 360^{\circ} = 210^{\circ}
\]
Теперь находим \( \sin 210^{\circ} \):
\[
\sin(210^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}
\]
Итак, значения тригонометрических функций:
\[
\sin 300^{\circ} = -\frac{\sqrt{3}}{2}, \quad \cos 330^{\circ} = \frac{\sqrt{3}}{2}, \quad \sin 570^{\circ} = -\frac{1}{2}
\]
Quick Answer
\[
\sin 300^{\circ} = -\frac{\sqrt{3}}{2}, \quad \cos 330^{\circ} = \frac{\sqrt{3}}{2}, \quad \sin 570^{\circ} = -\frac{1}{2}
\]
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