Giải bài 2 trang 153 – SGK môn Đại số lớp 10

a) Ta có: 
$\begin{align} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\ & \Leftrightarrow {{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+{{\cos }^{2}}\alpha =1 \\ & \Leftrightarrow {{\cos }^{2}}\alpha =\dfrac{2}{3} \\ & \Leftrightarrow \cos \alpha =\pm \dfrac{\sqrt{6}}{3} \\ \end{align}$
Do $0<\alpha <\dfrac{\pi }{2}$ nên $\cos \alpha >0$. Vậy $\cos \alpha =\dfrac{\sqrt{6}}{3}.$
Suy ra 
$\begin{align} & \cos \left( \alpha +\dfrac{\pi }{3} \right)=\cos \alpha \cos \dfrac{\pi }{3}-\sin \alpha \sin\dfrac{\pi }{3} \\ & =\dfrac{\sqrt{6}}{3}.\dfrac{1}{2}-\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{3}}{2} \\ & =\dfrac{\sqrt{6}-3}{6} \\ \end{align}$
b) Ta có: 
$\begin{align} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\ & \Leftrightarrow {{\sin }^{2}}\alpha +{{\left( -\dfrac{1}{3} \right)}^{2}}=1 \\ & \Leftrightarrow {{\sin }^{2}}\alpha =\dfrac{8}{9} \\ & \Leftrightarrow \sin \alpha =\pm \dfrac{2\sqrt{2}}{3} \\ \end{align}$
Do $\dfrac{\pi }{2}<\alpha <\pi$  nên $\sin \alpha >0$. Vậy $\sin \alpha =\dfrac{2\sqrt{2}}{3}\Rightarrow \tan \alpha =-2\sqrt{2}.$
Suy ra 
$\begin{align} & \tan \left( \alpha -\dfrac{\pi }{4} \right)=\dfrac{\tan \alpha -\tan \dfrac{\pi }{4}}{1+\tan \alpha .\tan \dfrac{\pi }{4}} \\ & =\dfrac{-2\sqrt{2}-1}{1-2\sqrt{2}} \\ & =\dfrac{9+4\sqrt{2}}{7} \\ \end{align}$
c) Ta có: 
$\begin{align} & {{\sin }^{2}}a+{{\cos }^{2}}a=1 \\ & \Leftrightarrow {{\left( \dfrac{4}{5} \right)}^{2}}+{{\cos }^{2}}a=1 \\ & \Leftrightarrow {{\cos }^{2}}a=\dfrac{9}{25} \\ & \Leftrightarrow \cos a=\pm \dfrac{3}{5} \\ \end{align}$
Do $0< a <{{90}^{o}}$ nên $\cos \alpha >0$. Vậy $\cos \,a=\dfrac{3}{5}.$
Tương  tự $\cos b=-\dfrac{\sqrt{5}}{3}$.
Suy ra 
$\begin{align} & +)\cos\left( a+b \right)=\cos a\cos b-\sin a\sin b \\ & =\dfrac{3}{5}.\left( -\dfrac{\sqrt{5}}{3} \right)-\dfrac{4}{5}.\dfrac{2}{3} \\ & =-\dfrac{3\sqrt{5}+8}{15} \\ \end{align}$
$\begin{align} +)&\,\,\sin\left( a-b \right)=\sin a\cos b-\cos a\operatorname{sinb} \\ & =\dfrac{4}{5}.\left( -\dfrac{\sqrt{5}}{3} \right)-\dfrac{3}{5}.\dfrac{2}{3} \\ & =-\dfrac{4\sqrt{5}+6}{15} \\ \end{align}$

Ghi nhớ:

$\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b;\\ \sin \left( a-b \right)=\sin a\cos b-\cos a\sin b;\\ \cos \left( a+b \right)=\cos a\cos b-\sin a\sin b;\\ \cos \left( a-b \right)=\cos a\cos b+\sin a\sin b;\\ \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a.\tan b}; \\ \tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a.\tan b}.$