گام به گام فعالیت صفحه 42 درس مثلثات ریاضی (1)

الف)

\(x = \sqrt {{4^2} + {3^2}}  = \sqrt {25}  = 5\)

\(\sin \theta  = \frac{{BC}}{{AC}} =\) \(\frac{3}{5}\)

\(\sin \alpha  =\) \(\frac{4}{5}\)

\(\cos \theta  =\) \(\frac{4}{5}   \)

\(\cos \alpha  =\) \(\frac{{BC}}{{AC}} = \frac{3}{5}\)

\(\tan \theta  = \frac{{BC}}{{AB}} = \frac{{\sin \theta }}{{\cos \theta }} =\) \(\frac{{\frac{3}{5}}}{{\frac{4}{5}}} = \frac{3}{4}\)

\(\tan \alpha  = \frac{{AB}}{{BC}} = \frac{{\sin \alpha }}{{\cos \alpha }} =\) \(\frac{{\frac{4}{5}}}{{\frac{3}{5}}} = \frac{4}{3}\)

\(\cot \theta  =\) \(\frac{1}{{\tan \theta }} = \frac{{\cos \theta }}{{\sin \theta }} = \frac{4}{3}\)

\(\cot \alpha  =\) \(\frac{1}{{\tan \alpha }} = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{3}{4}\)

ب)

\(\begin{array}{l}{\sin ^2}\;\theta  + {\cos ^2}\;\theta  = {\left( {\frac{3}{5}} \right)^2} + {\left( {\frac{4}{5}} \right)^2}\\ = \frac{9}{{25}} + \frac{{16}}{{25}} = \frac{{25}}{{25}} = 1\\\\{\sin ^2}\;\alpha  + {\cos ^2}\;\alpha  = {\left( {\frac{4}{5}} \right)^2} + {\left( {\frac{3}{5}} \right)^2}\\ = \frac{{16}}{{25}} + \frac{9}{{25}} = \frac{{25}}{{25}} = 1\end{array}\)

پ)

\(\begin{array}{l}{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = {\sin ^2}\theta + {\cos ^2}\theta \\\\ = {\left( {\frac{{BC}}{{AC}}} \right)^2} + {\left( {\frac{{AB}}{{AC}}} \right)^2} = \frac{{B{C^2} + A{B^2}}}{{A{C^2}}}\end{array}\)

\({ = \frac{{A{C^2}}}{{A{C^2}}} = 1}\)

ت)

\(\begin{array}{l}{\left( {\sin \;\alpha } \right)^2} + {\left( {\cos \;\alpha } \right)^2} = {\sin ^2}\;\alpha  + {\cos ^2}\;\alpha \\\\ = {\left( {\frac{{AB}}{{AC}}} \right)^2} + {\left( {\frac{{BC}}{{AC}}} \right)^2}\\\\ = \frac{{A{B^2} + B{C^2}}}{{A{C^2}}} = \frac{{A{C^2}}}{{A{C^2}}} = 1\end{array}\)