1. cho tan a =2 .tính sin 2a, cos2a,tan 2a 2. tính giá trị biểu thức: A=cos20 độ.cos40.cos60.cos80 B=sin10.sin50.sin70 15/10/2021 Bởi Rose 1. cho tan a =2 .tính sin 2a, cos2a,tan 2a 2. tính giá trị biểu thức: A=cos20 độ.cos40.cos60.cos80 B=sin10.sin50.sin70
Giải thích các bước giải: \(\begin{array}{l}1,\\\tan a = 2 \Leftrightarrow \dfrac{{\sin a}}{{\cos a}} = 2 \Leftrightarrow \sin a = 2\cos a\\{\sin ^2}a + {\cos ^2}a = 1\\ \Leftrightarrow {\left( {2\cos a} \right)^2} + {\cos ^2}a = 1\\ \Leftrightarrow {\cos ^2}a = \dfrac{1}{5}\\\sin 2a = 2\sin a.\cos a = 2.\cos a.\cos a = 2.{\cos ^2}a = \dfrac{2}{5}\\\cos 2a = 2{\cos ^2}a – 1 = 2.\dfrac{1}{5} – 1 = – \dfrac{3}{5}\\\tan 2a = \dfrac{{\sin 2a}}{{\cos 2a}} = – \dfrac{2}{3}\\2,\\A = \cos 20^\circ .\cos 40^\circ .\cos 60^\circ .\cos 80^\circ \\ \Leftrightarrow A.\sin 20^\circ = \sin 20^\circ .cos20^\circ .\cos 40^\circ .\dfrac{1}{2}.\cos 80^\circ \\ \Leftrightarrow A.\sin 20^\circ = \dfrac{1}{2}\sin 40^\circ .\cos 40^\circ .\dfrac{1}{2}.\cos 80^\circ \\ \Leftrightarrow A.\sin 20^\circ = \dfrac{1}{4}.\dfrac{1}{2}\sin 80^\circ .\cos 80^\circ \\ \Leftrightarrow A.\sin 20^\circ = \dfrac{1}{8}.\dfrac{1}{2}.sin160^\circ \\ \Leftrightarrow A.\sin 20^\circ = \dfrac{1}{{16}}.\sin \left( {180^\circ – 160^\circ } \right)\\ \Leftrightarrow A.\sin 20^\circ = \dfrac{1}{{16}}.\sin 20^\circ \\ \Leftrightarrow A = \dfrac{1}{{16}}\end{array}\) Bình luận
Giải thích các bước giải:
\(\begin{array}{l}
1,\\
\tan a = 2 \Leftrightarrow \dfrac{{\sin a}}{{\cos a}} = 2 \Leftrightarrow \sin a = 2\cos a\\
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( {2\cos a} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = \dfrac{1}{5}\\
\sin 2a = 2\sin a.\cos a = 2.\cos a.\cos a = 2.{\cos ^2}a = \dfrac{2}{5}\\
\cos 2a = 2{\cos ^2}a – 1 = 2.\dfrac{1}{5} – 1 = – \dfrac{3}{5}\\
\tan 2a = \dfrac{{\sin 2a}}{{\cos 2a}} = – \dfrac{2}{3}\\
2,\\
A = \cos 20^\circ .\cos 40^\circ .\cos 60^\circ .\cos 80^\circ \\
\Leftrightarrow A.\sin 20^\circ = \sin 20^\circ .cos20^\circ .\cos 40^\circ .\dfrac{1}{2}.\cos 80^\circ \\
\Leftrightarrow A.\sin 20^\circ = \dfrac{1}{2}\sin 40^\circ .\cos 40^\circ .\dfrac{1}{2}.\cos 80^\circ \\
\Leftrightarrow A.\sin 20^\circ = \dfrac{1}{4}.\dfrac{1}{2}\sin 80^\circ .\cos 80^\circ \\
\Leftrightarrow A.\sin 20^\circ = \dfrac{1}{8}.\dfrac{1}{2}.sin160^\circ \\
\Leftrightarrow A.\sin 20^\circ = \dfrac{1}{{16}}.\sin \left( {180^\circ – 160^\circ } \right)\\
\Leftrightarrow A.\sin 20^\circ = \dfrac{1}{{16}}.\sin 20^\circ \\
\Leftrightarrow A = \dfrac{1}{{16}}
\end{array}\)