شاه کلید نکات
شاه کلید سوال
حاصل هریک از نسبت های مثلثاتی زیر را مطابق نمونه بیابید.
\(\begin{array}{l}\sin {225^\circ } = \sin \left( {{{180}^\circ } + {{45}^\circ }} \right) = - \sin {45^\circ } = \frac{{ - \sqrt 2 }}{2}\\\tan {225^\circ } = ...........\\\cos \left( {\frac{{ - 4\pi }}{3}} \right) = \cos .... = \cos \left( {\pi + ....} \right) = ........... = ..........\\\sin \left( {\frac{{ - 7\pi }}{6}} \right) = .............\\\cot \left( {\frac{{5\pi }}{4}} \right) = .............\end{array}\)
\(\begin{array}{l}\sin {225^\circ } = \sin \left( {{{180}^\circ } + {{45}^\circ }} \right) = - \sin {45^\circ } = \frac{{ - \sqrt 2 }}{2}\\\tan {225^\circ } = \tan \left( {{{180}^ \circ } + {{45}^ \circ }} \right) = \tan {45^ \circ } = 1\\\cos \left( {\frac{{ - 4\pi }}{3}} \right) = \cos \frac{{4\pi }}{3} = \cos \left( {\pi + \frac{\pi }{3}} \right) = - \cos \frac{\pi }{3} = - \frac{1}{2}\\\sin \left( {\frac{{ - 7\pi }}{6}} \right) = - \sin \frac{{7\pi }}{6} = - \sin \left( {\pi + \frac{\pi }{6}} \right) = \sin \frac{\pi }{6} = \frac{1}{2}\\\cot \left( {\frac{{5\pi }}{4}} \right) = \cot \left( {\pi + \frac{\pi }{4}} \right) = \cot \frac{\pi }{4} = 1\end{array}\)
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