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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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: \[ \sin 300^{\circ} = -\frac{\sqrt{3}}{2}, \quad \cos 330^{\circ} = \frac{\sqrt{3}}{2}, \quad \sin 570^{\circ} = -\frac{1}{2} \]

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\[ \sin 300^{\circ} = -\frac{\sqrt{3}}{2}, \quad \cos 330^{\circ} = \frac{\sqrt{3}}{2}, \quad \sin 570^{\circ} = -\frac{1}{2} \]
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