شاه کلید نکات
شاه کلید سوال
مطابق نمونه هر یک از نسبت های مثلثاتی زاویه های زیر را مشخص کنید. (شکل سمت راست)
\(\begin{array}{l}\sin {750^\circ } = \sin \left( {2 \times {{360}^\circ } + 30} \right) = \sin {30^\circ } = \frac{1}{2}\\\tan \left( { - {{315}^\circ }} \right) = - \tan \left( {{{315}^\circ }} \right) = - \tan \left( {{{360}^\circ } - {{45}^\circ }} \right) = - \tan \left( {{{45}^\circ }} \right) = - \left( { - \tan {{45}^\circ }} \right) = \tan {45^\circ } = 1\end{array}\)
1 \(\cos {300^\circ } = .............\)
2 \(\sin {420^\circ } = .............\)
3 \(\tan \left( { - {{225}^\circ }} \right) = .............\)
4 \(\cot \left( { - {{330}^\circ }} \right) = .............\)
5 \(\sin \frac{{11\pi }}{4} = .............\)
6 \(\cos \left( {\frac{{ - 7\pi }}{4}} \right) = .............\)
1
\(\cos {300^\circ } = \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right) = \cos \left( { - {{60}^ \circ }} \right) = \cos {60^ \circ } = \frac{1}{2}\)
2
\(\sin {420^\circ } = \sin \left( {{{360}^ \circ } + {{60}^ \circ }} \right) = \sin {60^ \circ } = \frac{{\sqrt 3 }}{2}\)
3
\(\tan \left( { - {{225}^\circ }} \right) = - \tan {225^ \circ } = - \tan \left( {{{180}^ \circ } + {{45}^ \circ }} \right) = \tan {45^ \circ } = 1\)
4
\(\cot \left( { - {{330}^\circ }} \right) = - \cot {330^ \circ } = - \cot \left( {{{360}^ \circ } - {{30}^ \circ }} \right) = - \cot \left( { - {{30}^ \circ }} \right) = \cot {30^ \circ } = \sqrt 3 \)
5
\(\sin \frac{{11\pi }}{4} = \sin \left( {2\pi - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{6}} \right) = - \sin \frac{\pi }{6} = - \frac{1}{2}\)
6
\(\cos \left( {\frac{{ - 7\pi }}{4}} \right) = \cos \frac{{7\pi }}{4} = \cos \left( {2\pi - \frac{\pi }{4}} \right) = \cos \left( { - \frac{\pi }{4}} \right) = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}\)
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