Bài 5 trang 154 sách giáo khoa Đại Số 10

LG a

\(\displaystyle \sin a = -0,6\) và \(\displaystyle π < a < {{3\pi } \over 2}\)

Phương pháp giải:

+) Với \(\pi  < a < \frac{{3\pi }}{2}\) ta có \(\sin a < 0, \, \, \cos a < 0.\)

+) Với \(\frac{{\pi }}{2} < a < \pi\) ta có \(\sin a > 0, \, \, \cos a < 0.\)

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

+) \(\sin 2a = 2\sin a.\cos a.\)

+) \(\cos 2a = {\cos ^2}a - {\sin ^2}a \) \(= 2{\cos ^2}a - 1 \)\(= 1 - 2{\sin ^2}a.\)

Lời giải chi tiết:

\(\begin{array}{l}{\sin ^2}a + {\cos ^2}a = 1\\ \Rightarrow {\cos ^2}a = 1 - {\sin ^2}a\\ = 1 - {\left( { - 0,6} \right)^2} = 0,64\end{array}\)

Mà \(\pi  < a < \dfrac{{3\pi }}{2} \Rightarrow \cos a < 0\) \( \Rightarrow \cos a =  - \sqrt {0,64}  =  - 0,8\)

\( \Rightarrow \sin 2a = 2\sin a\cos a\) \( = 2.\left( { - 0,6} \right).\left( { - 0,8} \right) = \dfrac{{24}}{{25}}\)

\(\cos 2a = {\cos ^2}a - {\sin ^2}a\) \( = {\left( { - 0,8} \right)^2} - {\left( { - 0,6} \right)^2} = \dfrac{7}{{25}}\)

\( \Rightarrow \tan 2a = \dfrac{{\sin 2a}}{{\cos 2a}}\) \( = \dfrac{{24}}{{25}}:\dfrac{7}{{25}} = \dfrac{{24}}{7}\)

LG b

\(\displaystyle \cos a =  - {5 \over {13}}\) và \(\displaystyle {\pi  \over 2} < a < π\)

Lời giải chi tiết:

\(\begin{array}{l}{\sin ^2}a + {\cos ^2}a = 1\\ \Rightarrow {\sin ^2}a = 1 - {\cos ^2}a\\ = 1 - {\left( { - \dfrac{5}{{13}}} \right)^2} = \dfrac{{144}}{{169}}\end{array}\)

Mà \(\dfrac{\pi }{2} < a < \pi  \Rightarrow \sin a > 0\) \( \Rightarrow \sin a = \sqrt {\dfrac{{144}}{{169}}}  = \dfrac{{12}}{{13}}\)

\( \Rightarrow \sin 2a = 2\sin a\cos a\) \( = 2.\dfrac{{12}}{{13}}.\left( { - \dfrac{5}{{13}}} \right) =  - \dfrac{{120}}{{169}}\)

\(\cos 2a = 2{\cos ^2}a - 1\) \( = 2.{\left( { - \dfrac{5}{{13}}} \right)^2} - 1 =  - \dfrac{{119}}{{169}}\)

\(\tan 2a = \dfrac{{\sin 2a}}{{\cos 2a}}\) \( = \left( { - \dfrac{{120}}{{169}}} \right):\left( { - \dfrac{{119}}{{169}}} \right) = \dfrac{{120}}{{119}}\)

LG c

 \(\displaystyle \sin a + \cos a = {1 \over 2}\) và \(\dfrac{\pi }{2} < a < \dfrac{{3\pi }}{4}\)

Lời giải chi tiết:

\(\begin{array}{l}{\left( {\sin a + \cos a} \right)^2}\\ = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a\\ = 1 + \sin 2a\\ \Rightarrow \sin 2a = {\left( {\sin a + \cos a} \right)^2} - 1\\ = {\left( {\dfrac{1}{2}} \right)^2} - 1 =  - \dfrac{3}{4}\\ \Rightarrow \sin 2a =  - \dfrac{3}{4}\end{array}\)

Mà

\(\begin{array}{l}{\sin ^2}2a + {\cos ^2}2a = 1\\ \Rightarrow {\cos ^2}2a = 1 - {\sin ^2}2a\\ = 1 - {\left( { - \dfrac{3}{4}} \right)^2} = \dfrac{7}{{16}}\end{array}\)  

Lại có \(\dfrac{\pi }{2} < a < \dfrac{{3\pi }}{4}\) \( \Rightarrow \pi  < 2a < \dfrac{{3\pi }}{2}\) \( \Rightarrow \cos 2a < 0\)

\( \Rightarrow \cos 2a =  - \sqrt {\dfrac{7}{{16}}}  =  - \dfrac{{\sqrt 7 }}{4}\)

\( \Rightarrow \tan 2a = \dfrac{{\sin 2a}}{{\cos 2a}}\) \( = \left( { - \dfrac{3}{4}} \right):\left( { - \dfrac{{\sqrt 7 }}{4}} \right) = \dfrac{3}{{\sqrt 7 }}\)

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