شاه کلید نکات
شاه کلید سوال
1 سایر نسبت های مثلثاتی زاویهٔ \( - \frac{\pi }{4}\) رادیان را مطابق نمونه به دست آورید.
\(\sin ( - \frac{\pi }{4}) = - \sin \frac{\pi }{4}\)
\(\begin{array}{l}\sin ( - \frac{\pi }{4}) = - \sin \frac{\pi }{4} = - \frac{{\sqrt 2 }}{2}\\\\\cos ( - \frac{\pi }{4}) = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}\\\\\tan ( - \frac{\pi }{4}) = - \tan \frac{\pi }{4} = - 1\\\\\cot ( - \frac{\pi }{4}) = - \cot \frac{\pi }{4} = - 1\end{array}\)
2 حاصل هریک از عبارت های زیر را مطابق نمونه به دست آورید.
\(\begin{array}{l}\cot (\frac{{ - \pi }}{3}) \times \cos (\frac{{ - \pi }}{6}) + \tan (\frac{{ - \pi }}{4}) = \\\\ - \cot \frac{\pi }{3} \times \cos \frac{\pi }{6} - \tan \frac{\pi }{4} = \\\\ - \frac{{\sqrt 3 }}{3} \times \frac{{\sqrt 3 }}{2} - 1 = \frac{{ - 3}}{2}\end{array}\)
الف
\(\begin{array}{l}\frac{{\cos ( - {{90}^ \circ }) + \sin ( - {{270}^ \circ })}}{{\sin ( - {{180}^ \circ }) - \cos ( - {{360}^ \circ })}} = \\\\\frac{{..................}}{{..................}} = ........................\end{array}\)
ب
\(\cot (\frac{{ - \pi }}{6}) + \tan (\frac{{ - \pi }}{3}) = ..................\)
پ
\(\begin{array}{l}\cos ( - {45^ \circ }) \times \cos ( - {60^ \circ }) + \sin ( - {45^ \circ }) \times \sin ( - {60^ \circ }) = \\\\............... + ............... = ...............\end{array}\)
الف
\(\begin{array}{l}\frac{{\cos ( - {{90}^ \circ }) + \sin ( - {{270}^ \circ })}}{{\sin ( - {{180}^ \circ }) - \cos ( - {{360}^ \circ })}} = \\\\\frac{{\cos {{90}^ \circ } - \sin {{270}^ \circ }}}{{ - \sin {{180}^ \circ } - \cos {{360}^ \circ }}} = \frac{{0 + 1}}{{0 - 1}} = - 1\end{array}\)
ب
\(\begin{array}{l}\cot (\frac{{ - \pi }}{6}) + \tan (\frac{{ - \pi }}{3}) = - \cot \frac{\pi }{6} - \tan \frac{\pi }{3} = \\\\ - \sqrt 3 - \sqrt 3 = - 2\sqrt 3 \end{array}\)
پ
\(\begin{array}{l}\cos ( - {45^ \circ }) \times \cos ( - {60^ \circ }) + \sin ( - {45^ \circ }) \times \sin ( - {60^ \circ }) = \\\\\cos {45^ \circ } \times \cos {60^ \circ } - \sin {45^ \circ } \times ( - \sin {60^ \circ }) = \\\\\frac{{\sqrt 2 }}{2} \times \frac{1}{2} + \frac{{\sqrt 2 }}{2} \times \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 2 + \sqrt 6 }}{4}\end{array}\)
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