Cho tan a= $\frac{5}{9}$ Tính A=( 2sina -3cosa)^2 28/10/2021 Bởi Charlie Cho tan a= $\frac{5}{9}$ Tính A=( 2sina -3cosa)^2
Đáp án: \(A = \frac{{289}}{{106}}\) Giải thích các bước giải: \(\begin{array}{l}\tan a = \frac{5}{9}\\ \to \frac{{\sin a}}{{\cos a}} = \frac{5}{9}\\ \to \sin a = \frac{5}{9}\cos a\\Có:{\sin ^2}a + {\cos ^2}a = 1\\ \to {\left( {\frac{5}{9}\cos a} \right)^2} + {\cos ^2}a = 1\\ \to \frac{{25}}{{81}}{\cos ^2}a + {\cos ^2}a = 1\\ \to \frac{{106}}{{81}}{\cos ^2}a = 1\\ \to {\cos ^2}a = \frac{{81}}{{106}}\\A = {\left( {2\sin a – 3\cos a} \right)^2}\\ = {\left( {2.\frac{5}{9}\cos a – 3\cos a} \right)^2}\\ = {\left( { – \frac{{17}}{9}\cos a} \right)^2}\\ = \frac{{289}}{{81}}.{\cos ^2}a = \frac{{289}}{{81}}.\frac{{81}}{{106}} = \frac{{289}}{{106}}\end{array}\) Bình luận
Đáp án:
\(A = \frac{{289}}{{106}}\)
Giải thích các bước giải:
\(\begin{array}{l}
\tan a = \frac{5}{9}\\
\to \frac{{\sin a}}{{\cos a}} = \frac{5}{9}\\
\to \sin a = \frac{5}{9}\cos a\\
Có:{\sin ^2}a + {\cos ^2}a = 1\\
\to {\left( {\frac{5}{9}\cos a} \right)^2} + {\cos ^2}a = 1\\
\to \frac{{25}}{{81}}{\cos ^2}a + {\cos ^2}a = 1\\
\to \frac{{106}}{{81}}{\cos ^2}a = 1\\
\to {\cos ^2}a = \frac{{81}}{{106}}\\
A = {\left( {2\sin a – 3\cos a} \right)^2}\\
= {\left( {2.\frac{5}{9}\cos a – 3\cos a} \right)^2}\\
= {\left( { – \frac{{17}}{9}\cos a} \right)^2}\\
= \frac{{289}}{{81}}.{\cos ^2}a = \frac{{289}}{{81}}.\frac{{81}}{{106}} = \frac{{289}}{{106}}
\end{array}\)