Bài 1 trang 153 SGK Đại số 10

Câu hỏi: Tính

Câu a​

\(\cos {225^0}, \sin {240^0}, \cot( - {15^0}), \tan{75^0}\);
Phương pháp giải:
Áp dụng các công thức:
\(\begin{array}{l}
+ ) \cos \left({\alpha + {{180}^0}} \right) = - \cos \alpha .\\
+ ) \sin\left({\alpha + {{180}^0}} \right) = - \sin \alpha .\\
+ ) \cot \left({ - \alpha } \right) = - \cot \alpha .\\
+ ) \sin \left({\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\\
+ ) \cos \left({\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .\\
+ ) \tan\left({\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.
\end{array}\)
Lời giải chi tiết:
\(\cos{225^0} = \cos({180^0} +{45^0})\) \(= - \cos{45^{0}}\) \(= -\dfrac{\sqrt{2}}{2}\)
+) \(\sin{240^0} = \sin({180^0} +{60^0}) \)
\(= - \sin{60^0}=  -\dfrac{\sqrt{3}}{2}\)
+) \(\cot( - {15^0})= - \cot{15^0} \) \( =  - \cot \left( {{{90}^0} - {{75}^0}} \right)\)
\(=  - \tan{75^0} =- \tan({30^0} +{45^0})\)
\(=\dfrac{-\tan30^{0}-\tan45^{0}}{1-\tan30^{0}\tan45^{0}}\)
\(=\dfrac{-\dfrac{1}{\sqrt{3}}-1}{1-\dfrac{1}{\sqrt{3}}}\) \(=-\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\) \(=-\dfrac{(\sqrt{3}+1)^{2}}{2} \)
\(= -2 - \sqrt 3\)
+) \(\tan 75^0 = \tan \left( {{{90}^0} - {{15}^0}} \right)\) \(= \cot 15^0=-\cot (-15^0)\) \(=-(-2 - \sqrt 3)= 2 + \sqrt3\)

Câu b​

\(\sin \dfrac{7\pi}{12},\) \(\cos \left ( -\dfrac{\pi}{12} \right),\) \(\tan\left ( \dfrac{13\pi}{12} \right)\)
Lời giải chi tiết:
\(\sin \dfrac{7\pi}{12} = \sin \left ( \dfrac{\pi}{3}+\dfrac{\pi}{4} \right) \)
\(=\sin\dfrac{\pi }{3}\cos\dfrac{\pi}{4}+ \cos \dfrac{\pi }{3}\sin\dfrac{\pi}{4}\)
\(= \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} + \dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2} \) \(= \dfrac{{\sqrt 6 }}{4} + \dfrac{{\sqrt 2 }}{4} = \dfrac{{\sqrt 6  + \sqrt 2 }}{4}\)
+) \(\cos \left ( -\dfrac{\pi }{12} \right) = \cos \left (\dfrac{\pi }{4} -\dfrac{\pi }{3}\right) \)
\(= \cos \dfrac{\pi }{4}\cos\dfrac{\pi }{3} + \sin \dfrac{\pi }{3}\sin \dfrac{\pi }{4}\) \(=\dfrac{{\sqrt 2 }}{2} . \dfrac{1}{2}+ \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2} \) \(= \dfrac{{\sqrt 2 }}{4} + \dfrac{{\sqrt 6 }}{4} = \dfrac{{\sqrt 2  + \sqrt 6 }}{4}\)
+)  \(\tan \left ( \dfrac{13\pi }{12} \right) = \tan(π +  \dfrac{\pi }{12}) \)
\(= \tan \dfrac{\pi }{12} = \tan \left ( \dfrac{\pi }{3}-\dfrac{\pi}{4} \right)\)
\(= \dfrac{\tan\dfrac{\pi }{3}-\tan\dfrac{\pi }{4}}{1+\tan\dfrac{\pi }{3}\tan\dfrac{\pi }{4}}\)
\(= \dfrac{{\sqrt 3  - 1}}{{1 + \sqrt 3.1}} \) \(= \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}} \) \(= \dfrac{{{{\left( {\sqrt 3  - 1} \right)}^2}}}{{3 - 1}} \) \(= \dfrac{{4 - 2\sqrt 3 }}{2} = 2 - \sqrt 3 \)