Bài 1 trang 153 SGK Đại số 10

Đề bài

Tính

a) \(\cos {225^0},\, \sin {240^0}, \, \cot( - {15^0}), \, \tan{75^0}\);

b) \(\sin \frac{7\pi}{12},\) \(\cos \left ( -\frac{\pi}{12} \right ),\) \(\tan\left ( \frac{13\pi}{12} \right )\)

Hướng dẫn giải

Áp dụng các công thức:

\(\begin{array}{l}
+ )\;\cos \left( {\alpha + {{180}^0}} \right) = - \cos \alpha .\\
+ )\;sin\left( {\alpha + {{180}^0}} \right) = - \sin \alpha .\\
+ )\;\;\cot \left( { - \alpha } \right) = - \cot \alpha .\\
+ )\;\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\\
+ )\;\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .\\
+ )\;tan\left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.
\end{array}\)

Lời giải chi tiết

a) \(\cos{225^0} = \cos({180^0} +{45^0})= - \cos{45^{0}}\)

\(= -\frac{\sqrt{2}}{2}\)

+) \(\sin{240^0} = \sin({180^0} +{60^0}) \)

 \(= - \sin{60^0}=  -\frac{\sqrt{3}}{2}\)

+) \(\cot( - {15^0})= - \cot{15^0} \)

\(=  - \tan{75^0} =- \tan({30^0} +{45^0})\)

\( =\frac{-\tan30^{0}-\tan45^{0}}{1-\tan30^{0}\tan45^{0}}\)

\(=\frac{-\frac{1}{\sqrt{3}}-1}{1-\frac{1}{\sqrt{3}}}=-\frac{\sqrt{3}+1}{\sqrt{3}-1}=-\frac{(\sqrt{3}+1)^{2}}{2} \)

\(= -2 - \sqrt 3\)

+) \(\tan 75^0= \cot15^0= 2 + \sqrt3\)

b) \(\sin \frac{7\pi}{12} = \sin \left ( \frac{\pi}{3}+\frac{\pi}{4} \right ) \)

\(=\sin\frac{\pi }{3}\cos\frac{\pi}{4}+ \cos \frac{\pi }{3}\sin\frac{\pi}{4}\)

\( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )=\frac{\sqrt{6}+\sqrt{2}}{4}\)

+) \(\cos \left ( -\frac{\pi }{12} \right ) = \cos \left ( \frac{\pi }{4} -\frac{\pi }{3}\right ) \)

                    \(= \cos \frac{\pi }{4}\cos\frac{\pi }{3} + sin \frac{\pi }{3}sin \frac{\pi }{4}\) 

                    \( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )=0,9659\)

+)  \(\tan \left ( \frac{13\pi }{12} \right ) = \tan(π +  \frac{\pi }{12}) \)

\(= \tan \frac{\pi }{12} = \tan \left ( \frac{\pi }{3}-\frac{\pi}{4} \right )\)

\(= \frac{\tan\frac{\pi }{3}-\tan\frac{\pi }{4}}{1+\tan\frac{\pi }{3}\tan\frac{\pi }{4}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}= 2 - \sqrt3\)