Bài 32 trang 206 SGK Đại số 10 Nâng cao

LG a

\(\sin \alpha  = {4 \over 5}\,\,;\,\,\,\cos \alpha  < 0\)

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha
\end{array}\)

\(\eqalign{
&\cos \alpha <0 \Rightarrow \cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } \cr &= - \sqrt {1 - {{16} \over {25}}} = - {3 \over 5} \cr 
& \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = - {4 \over 3} \cr 
& \cot \alpha = {1 \over {\tan \alpha }} = - {3 \over 4} \cr} \)

LG b

\(\cos \alpha  =  - {8 \over {17}};\,\,\,{\pi  \over 2} < \alpha  < \pi \)

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Rightarrow {\sin ^2}\alpha = 1 - {\cos ^2}\alpha
\end{array}\)

\(\eqalign{
& \,{\pi \over 2} < \alpha < \pi \Rightarrow \sin \alpha > 0 \cr 
& \Rightarrow \sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } \cr &= \sqrt {1 - {{({-8 \over {17}})}^2}} = {{15} \over {17}} \cr 
& \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = - {{15} \over 8} \cr 
& \cot \alpha = {1 \over {\tan \alpha }} = - {8 \over {15}} \cr} \) 

LG c

\(\tan \alpha  = \sqrt 3 \,\,;\,\,\,\pi  < \alpha  < {{3\pi } \over 2}\)

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}
1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\\
\Rightarrow {\cos ^2}\alpha = \frac{1}{{1 + {{\tan }^2}\alpha }}
\end{array}\)

\(\eqalign{
& \pi < \alpha < {{3\pi } \over 2} \Rightarrow \cos \alpha < 0 \cr 
& \Rightarrow \cos \alpha = {{ - 1} \over {\sqrt {1 + {{\tan }^2}\alpha } }}\cr & = {{ - 1} \over {\sqrt {1 + {{(\sqrt 3 )}^2}} }} = - {1 \over 2} \cr 
& \tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} \cr &\Rightarrow \sin \alpha  = \tan \alpha \cos \alpha  \cr &= \sqrt 3 .\left( { - \frac{1}{2}} \right)= - {{\sqrt 3 } \over 2} \cr 
& \cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{{\sqrt 3 }} = {{\sqrt 3 } \over 3} \cr} \)

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