Giải câu 1 bài 3: Công thức lượng giác sgk Đại số 10 trang 153.
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.
- \(\sin \frac{7\pi}{12} = \sin \left ( \frac{\pi}{3}+\frac{\pi}{4} \right ) \)
\(=\sin\frac{\pi }{3}\cos\frac{\pi}{4}+ \cos \frac{\pi }{3}\sin\frac{\pi}{4}\)
\( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )\)
\(=\frac{\sqrt{6}+\sqrt{2}}{4}\)
- \(\cos \left ( -\frac{\pi }{12} \right ) = \cos \left ( \frac{\pi }{4} -\frac{\pi }{3}\right ) \)
\(= \cos \frac{\pi }{4}\cos\frac{\pi }{3} + sin \frac{\pi }{3}sin \frac{\pi }{4}\)
\( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )\)
\(=\frac{\sqrt{6}+\sqrt{2}}{4}\)
- \(\tan \left ( \frac{13\pi }{12} \right ) = \tan(π + \frac{\pi }{12}) = \tan \frac{\pi }{12} = \tan \left ( \frac{\pi }{3}-\frac{\pi}{4} \right )\)
\(= \frac{\tan\frac{\pi }{3}-\tan\frac{\pi }{4}}{1+\tan\frac{\pi }{3}\tan\frac{\pi }{4}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}= 2 - \sqrt3\)