a) Cho cos alpha = 4/13 thoả mãn 0 < alpha < pi/ 2 . Tính sin alpha ? b) Cho tan alpha = - 2. Tính A = sin^2 alpha - 2 sin alpha cos alpha + 1 / sin ^2 alpha - sin al[ha cos alpha + 3

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a) Cho \[\cos \alpha = \frac{4}{{13}}\] thoả mãn\[0 < \alpha < \frac{\pi }{2}\]. Tính \[\sin \alpha \]?

b) Cho\(\tan \alpha = - 2\). Tính \(A = \frac{{{{\sin }^2}\alpha - 2\sin \alpha \cos \alpha + 1}}{{{{\sin }^2}\alpha - \sin \alpha \cos \alpha + 3{{\cos }^2}\alpha }}\).

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

a) Vì \[0 < \alpha < \frac{\pi }{2}\] nên \[\sin \alpha > 0\].

Ta có \[{\sin ^2}\alpha + {\cos ^2}\alpha = 1\]nên:

\({\left( {\frac{4}{{13}}} \right)^2}\) + sin²a = 1 Û sin²a = 1 - \(\frac{{16}}{{169}}\) Û sina = \(\frac{{3\sqrt {17} }}{{13}}\).

b) \(\tan \alpha = - 2\)

\(\begin{array}{l}A = \frac{{{{\sin }^2}\alpha - 2\sin \alpha \cos \alpha + 1}}{{{{\sin }^2}\alpha - \sin \alpha \cos \alpha + 3{{\cos }^2}\alpha }}\\\,\,\,\, = \frac{{2{{\tan }^2}\alpha - 2\tan \alpha + 1}}{{{{\tan }^2}\alpha - \tan \alpha + 3}} = \frac{{13}}{9}\end{array}\)