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Теория: 05 Тригонометрические функции

Задание

Тригонометрия

Правило

Основные тригонометрические формулы

Основное тригонометрическое тождество

\(\displaystyle \sin^2\color{blue}{\alpha}+\cos^2\color{blue}{\alpha}=1\)

    Формулы двойного угла

    • \(\displaystyle \sin2\color{blue}{\alpha}=2\sin\color{blue}{\alpha}\cos\color{blue}{\alpha}\)
    • \(\displaystyle \cos2\color{blue}{\alpha}=\cos^2\color{blue}{\alpha}-\sin^2\color{blue}{\alpha}\)
    • \(\displaystyle \cos2\color{blue}{\alpha}=1-2\sin^2\color{blue}{\alpha}\)
    • \(\displaystyle \cos2\color{blue}{\alpha}=2\cos^2\color{blue}{\alpha}-1\)

    Тангенс и котангенс

    • \(\displaystyle \tg\alpha=\frac{\sin\alpha}{\cos\alpha}\)
    • \(\displaystyle \ctg\alpha=\frac{\cos\alpha}{\sin\alpha}\)
    • \(\displaystyle \tg\color{blue}{\alpha}\cdot\ctg\color{blue}{\alpha}=1\)


    Таблица основных значений тригонометрических функций

    \(\displaystyle \alpha\)\(\displaystyle \frac{\pi}{6}(30^{\circ})\)\(\displaystyle \frac{\pi}{4}(45^{\circ})\)\(\displaystyle \frac{\pi}{3}(60^{\circ})\)
    \(\displaystyle \sin \alpha\)\(\displaystyle \frac{1}{2}\)\(\displaystyle \frac{\sqrt{2}}{2}\)\(\displaystyle \frac{\sqrt{3}}{2}\)
    \(\displaystyle \cos \alpha\)\(\displaystyle \frac{\sqrt{3}}{2}\)\(\displaystyle \frac{\sqrt{2}}{2}\)\(\displaystyle \frac{1}{2}\)
    \(\displaystyle \tg \alpha\)\(\displaystyle \frac{\sqrt{3}}{3}\)\(\displaystyle 1\)\(\displaystyle \sqrt{3}\)
    \(\displaystyle \ctg \alpha\)\(\displaystyle \sqrt{3}\)\(\displaystyle 1\)\(\displaystyle \frac{\sqrt{3}}{3}\)

    Решение