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

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

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