Bài 2 trang 154 SGK Đại số 10

LG a

$\cos(α +  \dfrac{\pi}{3}),$ biết $\sinα =  \dfrac{1}{\sqrt{3}}$ và $0 < α <  \dfrac{\pi }{2}.$

Phương pháp giải:

+) Với $0 < \alpha  < \dfrac{\pi }{2}$ ta có: $\sin \alpha >0, \, \, \cos \alpha >0.$

+) Với $\dfrac{\pi }{2} < \alpha  < \pi$ ta có: $\sin \alpha >0, \, \, \cos \alpha < 0.$

+) $\sin^2 \alpha +\cos^2 \alpha =1.$

+) ${\tan ^2}x + 1 = \dfrac{1}{{{{\cos }^2}x}}.$

+)  ${\cot ^2}x + 1 = \dfrac{1}{{{{\sin }^2}x}}.$

+) $\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .$

+) $\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .$

+) $\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta .$

+) $\tan\left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.$

+) $\tan\left( {\alpha - \beta } \right) = \dfrac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}.$

Lời giải chi tiết:

Ta có:

$\begin{array}{l}{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\ \Rightarrow {\cos ^2}\alpha  = 1 - {\sin ^2}\alpha \\ = 1 - {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} = \dfrac{2}{3}\end{array}$

Mà $0 < \alpha  < \dfrac{\pi }{2} \Rightarrow \cos \alpha  > 0$

$\Rightarrow \cos \alpha  = \sqrt {\dfrac{2}{3}}  = \dfrac{{\sqrt 6 }}{3}$

$\Rightarrow \cos(α + \dfrac{\pi}{3}) = \cosα\cos \dfrac{\pi }{3} - \sinα\sin \dfrac{\pi}{3}$

$=  \dfrac{\sqrt{6}}{3}.\dfrac{1}{2}-\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{6}-3}{6}$

LG b

$\tan(α -   \dfrac{\pi }{4}),$ biết $\cosα = -\dfrac{1}{3}$ và $\dfrac{\pi }{2} < α < π.$

Lời giải chi tiết:

$\begin{array}{l}{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\ \Rightarrow {\sin ^2}\alpha  = 1 - {\cos ^2}\alpha \\ = 1 - {\left( { - \dfrac{1}{3}} \right)^2} = \dfrac{8}{9}\end{array}$

Mà $\dfrac{\pi }{2} < \alpha  < \pi  \Rightarrow \sin \alpha  > 0$

$\Rightarrow \sin \alpha  = \sqrt {\dfrac{8}{9}}  = \dfrac{{2\sqrt 2 }}{3}$

$\Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }}$ $= \dfrac{{2\sqrt 2 }}{3}:\left( { - \dfrac{1}{3}} \right) =  - 2\sqrt 2$

$\tan(α -  \dfrac{\pi}{4})$ $=  \dfrac{\tan\alpha -\tan\dfrac{\pi}{4}}{1+\tan\alpha \tan\dfrac{\pi}{4}}$ $=\dfrac{-1-2\sqrt{2}}{1-2\sqrt{2}}$ $=\dfrac{2\sqrt{2}+1}{2\sqrt{2}-1}$ $= \dfrac{{{{\left( {2\sqrt 2  + 1} \right)}^2}}}{{8 - 1}}$ $= \dfrac{{9 + 4\sqrt 2 }}{7}.$

LG c

$\cos(a + b), \, \, \sin(a - b)$ biết $\sin a =  \frac{4}{5}$ $0^0< a < 90^0,$ và $\sin b =  \frac{2}{3},$ $90^0< b < 180^0.$ 

Lời giải chi tiết:

$\begin{array}{l}{\sin ^2}a + {\cos ^2}a = 1\\ \Rightarrow {\cos ^2}a = 1 - {\sin ^2}a\\ = 1 - {\left( {\dfrac{4}{5}} \right)^2} = \dfrac{9}{{25}}\end{array}$

Mà ${0^0} < a < {90^0} \Rightarrow \cos a > 0$ $\Rightarrow \cos a = \sqrt {\dfrac{9}{{25}}}  = \dfrac{3}{5}$.

$\begin{array}{l}{\sin ^2}b + {\cos ^2}b = 1\\ \Rightarrow {\cos ^2}b = 1 - {\sin ^2}b\\ = 1 - {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{5}{9}\end{array}$

Mà ${90^0} < b < {180^0} \Rightarrow \cos b < 0$ $\Rightarrow \cos b =  - \sqrt {\dfrac{5}{9}}  =  - \dfrac{{\sqrt 5 }}{3}$.

$\cos(a + b) = \cos a\cos b - \sin a\sin b$

$=\dfrac{3}{5}\left ( -\dfrac{\sqrt{5}}{3} \right )-\dfrac{4}{5}.\dfrac{2}{3}=-\dfrac{3\sqrt{5}+8}{15}$