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Вычисли \(\cos \alpha\) и \(\sin \alpha\) , если:

1. \(\tg \alpha = 3\) .
2. \(\tg \alpha = - \dfrac{3}{4}\) .

Решение.

1. Выразим \(\cos^2 \alpha\) через \(\tg^2 \alpha\) : \(\newline\) \(1 + \tg^2 \alpha = \dfrac{1}{\cos^2 \alpha}\) , \(\newline\) \(\raisebox{-1em}{\) \cos^2 \alpha=\(}\) \(\quad\;\; \raisebox{-0.2em}{\)1\(} \qquad\) \(\raisebox{-0.7em}{\)\mathllap{\underline{\kern{4em}};}\(}\) \(\raisebox{-1em}{\)=\(}\) \(\quad\kern{0.5em}\) \(\raisebox{-0.2em}{\)1\(}\) \(\qquad\) \(\raisebox{-0.7em}{\)\mathllap{\underline{\kern{4em}};}\(}\) \(\raisebox{-1em}{\)=\(}\) \(\newline\) \(\kern{3.6em}\) \( 1 + \tg^2 \alpha\) \(\kern{1.5em}1+\) [ ]

\(=\) [ ]. \(\cos \alpha \) [ ] \(0\) , так как \(0 \degree \lt \alpha \lt 90 \degree\) ( по [ ]). Следовательно, \(\cos \alpha =\) [ ]. Вычислим \(\sin \alpha \cdot \tg \alpha =\dfrac{\sin \alpha}{\cos \alpha}\) , значит, \(\sin \alpha =\) [ ] \(=\) [ ] \(=\) [ ].

2. \(\cos^2 \alpha = \dfrac{1}{1 + \tg^2 \alpha} = \) [ ] \(=\) [ ]. Отметим, что \(\cos \alpha\) [ ] \(0\) , так как \(90 \degree \lt \alpha \lt 180 \degree\) ( по условию \(\tg \alpha \lt 0\) ). Следовательно, \(\cos \alpha =\) [ ]. \(\sin \alpha =\) [ ] \(=\) [ ] \(=\) [ ].

Ответ:

1. \(\cos \alpha = \) [ ], \(\sin \alpha =\) [ ];

2. \(\cos \alpha = \) [ ], \(\sin \alpha =\) [ ].