شاه کلید نکات
شاه کلید سوال
1 اگر \(\cos x = \frac{{ - 4}}{5}\) و sinx>0، نسبت های مثلثاتی دیگر زاویهٔ x را بیابید.
\(\begin{array}{l}\left. \begin{array}{l}\cos x = \frac{{ - 4}}{5} < 0\\\sin x > 0\end{array} \right\} \Rightarrow \\{\sin ^2}x = 1 - {\cos ^2}x = \frac{9}{{25}}\mathop \to \limits^{\sin x > 0} \mathop {}\nolimits^{} \sin x = \frac{3}{5}\\\tan x = \frac{{\sin x}}{{\cos x}} = - \frac{3}{4}\\\cot x = \frac{1}{{\tan x}} = - \frac{4}{3}\end{array}\)
2 جدول زیر را کامل کنید.
3 حاصل عبارت های زیر را به دست آورید.
الف \(\cot \frac{\pi }{6} - \tan \frac{\pi }{3} \times \sin \frac{\pi }{4} = \)
ب \(\frac{{{{\tan }^2}\left( {\frac{\pi }{6}} \right) + {{\sin }^2}\left( {\frac{\pi }{4}} \right)}}{{{{\cot }^2}\left( {\frac{\pi }{4}} \right) - {{\cos }^2}\left( {\frac{\pi }{3}} \right)}} + {\cos ^2}{75پ^\circ } + {\sin ^2}{75^\circ } = \)
الف
\(\begin{array}{l}\cot \frac{\pi }{6} - \tan \frac{\pi }{3} \times \sin \frac{\pi }{4} = \\\\\sqrt 3 - \sqrt 3 \times \frac{{\sqrt 2 }}{2} = \sqrt 3 - \frac{{\sqrt 6 }}{2} = \frac{{2\sqrt 3 - \sqrt 6 }}{2}\end{array}\)
ب
\(\begin{array}{l}\frac{{{{\tan }^2}\left( {\frac{\pi }{6}} \right) + {{\sin }^2}\left( {\frac{\pi }{4}} \right)}}{{{{\cot }^2}\left( {\frac{\pi }{4}} \right) - {{\cos }^2}\left( {\frac{\pi }{3}} \right)}} + {\cos ^2}{75^\circ } + {\sin ^2}{75^\circ } = \\\\\frac{{{{\left( {\frac{{\sqrt 3 }}{3}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}}{{{1^2} - {{\left( {\frac{1}{2}} \right)}^2}}} + 1 = \frac{{\frac{1}{3} + \frac{1}{2}}}{{1 - \frac{1}{4}}} + 1 = \frac{{19}}{9}\end{array}\)
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