Nếu biết \(\tan a = \dfrac{1}{2}\,\,\left( {0 <...

A

- Hướng dẫn giải

Phương pháp giải:

+) Sử dụng công thức \(\cos \left( {2a - b} \right) = \cos 2a\cos b + \sin 2a\sin b\).

+) Tính \(\sin 2a,\,\,\cos 2a,\,\,\sin b,\,\,\cos b\).

\(\cos \left( {2a - b} \right) = \cos 2a\cos b + \sin 2a\sin b\).

+) Ta có: \(\tan a = \dfrac{1}{2} \Rightarrow \dfrac{1}{{{{\cos }^2}a}} = 1 + {\tan ^2}a = 1 + \dfrac{1}{4} = \dfrac{5}{4}\).

Giải chi tiết:

\(\cos \left( {2a - b} \right) = \cos 2a\cos b + \sin 2a\sin b\).

+) Ta có: \(\tan a = \dfrac{1}{2} \Rightarrow \dfrac{1}{{{{\cos }^2}a}} = 1 + {\tan ^2}a = 1 + \dfrac{1}{4} = \dfrac{5}{4}\).

\(\Rightarrow {{\cos }^{2}}a=\frac{4}{5}\Leftrightarrow \cos a=\pm \frac{2}{\sqrt{5}}\).

Mà \(0<a<{{90}^{0}}\Rightarrow \cos a>0\Rightarrow \cos a=\frac{2}{\sqrt{5}}\).

\(\Rightarrow \sin a=\tan a.\cos a=\frac{1}{2}.\frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}\)

\( \Rightarrow \left\{ \matrix{
\sin 2a = 2\sin a\cos a = 2.{1 \over {\sqrt 5 }}.{2 \over {\sqrt 5 }} = {4 \over 5} \hfill \cr
\cos 2a = 1 - 2{\sin ^2}a = 1 - 2.{\left( {{1 \over {\sqrt 5 }}} \right)^2} = {3 \over 5} \hfill \cr} \right.\)

+) Ta có: \(\tan b =  - {1 \over 3} \Rightarrow {1 \over {{{\cos }^2}b}} = 1 + {\tan ^2}b = 1 + {1 \over 9} = {{10} \over 9}\)

\( \Rightarrow {\cos ^2}b = \dfrac{9}{{10}} \Leftrightarrow \cos b =  \pm \dfrac{3}{{\sqrt {10} }}\). Mà \({90^0} < b < {180^0} \Rightarrow \cos b < 0 \Leftrightarrow \cos b =  - \dfrac{3}{{\sqrt {10} }}\).

\(\Rightarrow \sin b=\tan b.\cos b=\dfrac{-1}{3}.\frac{-3}{\sqrt{10}}=\dfrac{1}{\sqrt{10}}\).

Vậy \(\cos \left( 2a-b \right)=\cos 2a\cos b+\sin 2a\sin b=\dfrac{3}{5}.\dfrac{-3}{\sqrt{10}}+\dfrac{4}{5}.\dfrac{1}{\sqrt{10}}=\dfrac{-1}{\sqrt{10}}=\dfrac{-\sqrt{10}}{10}\).

Chọn A.