گام به گام تمرین صفحه 45 درس 2 ریاضی (1) (مثلثات)

1)

ربع دوم   \(\Rightarrow \left\{ \begin{array}{l}\cos \alpha  < 0\\\sin \,\,\alpha  > 0\end{array} \right.\)

\(\begin{array}{l} \Rightarrow \sin \,\,\alpha  = \sqrt {1 - {{\cos }^2}\alpha }  = \sqrt {1 - \frac{9}{{25}}}  = \sqrt {\frac{{16}}{{25}}}  = \frac{4}{5}\\\tan \alpha  = \frac{{\sin \,\,\alpha }}{{\cos \alpha }} =  - \frac{4}{3}\\\cot \alpha  = \frac{{\cos \alpha }}{{\sin \,\,\alpha }} =  - \frac{3}{4}\end{array}\)

2)

ربع چهارم  \( \Rightarrow \left\{ \begin{array}{l}\cos \alpha > 0\\\sin \,\,\alpha < 0\end{array} \right. \)

\(\begin{array}{l}1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow {\cos ^2}\alpha = \frac{1}{{1 + {{\tan }^2}\alpha }} = \frac{9}{{25}}\\ \Rightarrow \cos \alpha = \frac{3}{5}\\\sin \;\alpha = - \sqrt {1 - {{\cos }^2}\alpha } = - \sqrt {\frac{{16}}{{25}}} = - \frac{4}{5}\\\cot \alpha = \frac{1}{{\tan \alpha }} = - \frac{3}{4}\end{array}\)

3)

\({90^ \circ } < {135^ \circ } < {180^ \circ } \Rightarrow\) ربع دوم  \(\Rightarrow \left\{ \begin{array}{l}\cos {135^ \circ } > 0\\\sin \,\,{135^ \circ } < 0\end{array} \right.\)

\(\begin{array}{l}\cos {135^ \circ } =  - \sqrt {1 - {{\sin }^2}{{135}^ \circ }}  =  - \sqrt {1 - \frac{1}{2}}  =  - \frac{{\sqrt 2 }}{2}\\\tan {135^ \circ } = \frac{{\sin \,\,{{135}^ \circ }}}{{\cos {{135}^ \circ }}} =  - 1\\\cot \alpha  = \frac{{\cos {{135}^ \circ }}}{{\sin \,\,{{135}^ \circ }}} =  - 1\end{array}\)

4)

\({180^ \circ } < {240^ \circ } < {270^ \circ } \Rightarrow\) ربع چهارم  \(\Rightarrow \left\{ \begin{array}{l}\cos {240^ \circ } < 0\\\sin \,\,{240^ \circ } < 0\end{array} \right.\)

\(\begin{array}{l}{180^ \circ } < {240^ \circ } < {270^ \circ } \Rightarrow \begin{array}{*{20}{c}}{}&{}&{}\end{array} \Rightarrow \left\{ \begin{array}{l}\cos {240^ \circ } < 0\\\sin \,\,{240^ \circ } < 0\end{array} \right.\\1 + {\tan ^2}{240^ \circ } = \frac{1}{{{{\cos }^2}{{240}^ \circ }}} \Rightarrow \\{\cos ^2}{240^ \circ } = \frac{1}{{1 + {{\tan }^2}{{240}^ \circ }}} = \frac{1}{4} \Rightarrow \\\cos {240^ \circ } =  - \frac{1}{2}\\\sin \;{240^ \circ } =  - \sqrt {1 - {{\cos }^2}{{240}^ \circ }}  =  - \sqrt {1 - \frac{1}{4}}  =  - \frac{{\sqrt 3 }}{2}\\\cot {240^ \circ } = \frac{1}{{\tan {{240}^ \circ }}} = \frac{1}{{\sqrt 3 }} = \frac{{\sqrt 3 }}{3}\end{array}\)

5)

\(\begin{array}{l}\sin \;{20^ \circ } \simeq 0/34 \Rightarrow \\\cos \;{20^ \circ } = \sqrt {1 - {{\sin }^2}\;{{20}^ \circ }}  = \sqrt {1 - {{\left( {0/34} \right)}^2}}  \simeq 0/94\\\tan \;{20^ \circ } = \frac{{AC}}{{BC}} = \frac{{\sin \;{{20}^ \circ }}}{{\cos \;{{20}^ \circ }}} \simeq 0/36\\AC = BC \times \tan \;{20^ \circ } = 200 \times 0/36 = 72\end{array}\)

6)

الف) \(\frac{1}{{\sin \theta }} \times \tan \theta  = \frac{1}{{\cos \theta }}\)

: طرف چپ  \(\frac{1}{{\sin \theta }} \times \tan \theta  = \frac{1}{{\sin \theta }} \times \frac{{\sin \theta }}{{\cos \theta }} = \frac{1}{{\cos \theta }}\) : طرف راست

ب) \(\frac{{\cos \theta }}{{1 + \sin \theta }} = \frac{{1 - \sin \theta }}{{\cos \theta }}\)

: طرف چپ  \(\frac{{\cos \theta }}{{1 + \sin \theta }} = \) 

\(\frac{{\cos \theta }}{{1 + \sin \theta }} \times \frac{{1 - \sin \theta }}{{1 - \sin \theta }} = \frac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{1 - {{\sin }^2}\theta }} = \)

\(\frac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{{{\cos }^2}\theta }} = \)

\(\frac{{1 - \sin \theta }}{{\cos \theta }}\)  : طرف راست

پ) \(\frac{{1 + \tan \alpha }}{{1 + \cot \alpha }} = \tan \alpha\)

طرف چپ :  \(\frac{{1 + \tan \alpha }}{{1 + \cot \alpha }} = \frac{{1 + \frac{{\sin \alpha }}{{\cos \alpha }}}}{{1 + \frac{{\cos \alpha }}{{\sin \alpha }}}} = \frac{{\frac{{\cos \alpha  + \sin \alpha }}{{\cos \alpha }}}}{{\frac{{\sin \alpha  + \cos \alpha }}{{sin\alpha }}}} = \frac{{sin\alpha }}{{\cos \alpha }} = \tan \alpha\) : طرف راست

ت) \(1 - \frac{{{{\cos }^2}x}}{{1 + \sin x}} = \sin x\)

: طرف چپ  \(1 - \frac{{{{\cos }^2}\;x}}{{1 + \sin \;x}} = \)

\(1 - \frac{{1 - si{n^2}\;x}}{{1 + \sin \;x}} = 1 - \frac{{\left( {1 + sin\;x} \right)\left( {1 - sin\;x} \right)}}{{1 + \sin \;x}} = \)

\({1 - \left( {1 - sin\;x} \right) = sin\;x}\)  : طرف راست

ث) \(\frac{1}{{\cos x}} - \tan x = \frac{{\cos x}}{{1 + \sin x}}\)

: طرف چپ  \(\frac{1}{{\cos \;x}} - \tan \;x = \)

\(\begin{array}{*{20}{l}}{\frac{1}{{\cos \;x}} - \frac{{\sin \;x}}{{\cos \;x}} = \frac{{1 - \sin \;x}}{{\cos \;x}} = }\\\begin{array}{l}\frac{{1 - \sin \;x}}{{\cos \;x}} \times \frac{{1 + \sin \;x}}{{1 + \sin \;x}} = \\\frac{{1 - {{\sin }^2}\;x}}{{\cos \;x\left( {1 + \sin \;x} \right)}} = \frac{{{{\cos }^2}\;x}}{{\cos \;x\left( {1 + \sin \;x} \right)}}\end{array}{}\end{array}\)

\({ = \frac{{\cos \;x}}{{1 + \sin \;x}}} \)  : طرف راست