جواب تمرین صفحه 112 درس 4 حسابان یازدهم (مثلثات)

الف

\(\begin{array}{l}\cos {15^\circ } = \cos \left( {{{60}^ \circ } - {{45}^ \circ }} \right)\\\\ = \cos {60^ \circ }\cos {45^ \circ } + \sin {60^ \circ }\sin {45^ \circ }\\\\ = \frac{1}{2} \times \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 + \sqrt 6 }}{4}\end{array}\)

ب

\(\tan {105^\circ } = \tan \left( {{{60}^ \circ } + {{45}^ \circ }} \right) = \frac{{\sin \left( {{{60}^ \circ } + {{45}^ \circ }} \right)}}{{\cos \left( {{{60}^ \circ } + {{45}^ \circ }} \right)}}\)

\(\begin{array}{l} = \frac{{\sin {{60}^ \circ }\cos {{45}^ \circ } + \cos {{60}^ \circ }\sin {{45}^ \circ }}}{{\cos {{60}^ \circ }\cos {{45}^ \circ } - \sin {{60}^ \circ }\sin {{45}^ \circ }}}\\\\ = \frac{{\frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2} + \frac{1}{2} \times \frac{{\sqrt 2 }}{2}}}{{\frac{1}{2} \times \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2}}} = \frac{{\frac{{\sqrt 6 + \sqrt 2 }}{4}}}{{\frac{{\sqrt 2 - \sqrt 6 }}{4}}} = \frac{{\sqrt 2 + \sqrt 6 }}{{\sqrt 2 - \sqrt 6 }} = - 2 - \sqrt 3 \end{array}\)

پ

\(\begin{array}{l}\sin \frac{\pi }{{12}} = \sin \left( {\frac{\pi }{3} - \frac{\pi }{4}} \right) = \sin \frac{\pi }{3}\cos \frac{\pi }{4} - \cos \frac{\pi }{3}\sin \frac{\pi }{4}\\\\ = \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2} - \frac{1}{2} \times \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 - \sqrt 2 }}{4}\end{array}\)

ت

\(\begin{array}{l}\sin {120^\circ } = \sin ({180^\circ } - {60^\circ }) = \\\\ - \sin {60^\circ } = - \frac{{\sqrt 3 }}{2}\end{array}\)

ث

\(\begin{array}{l}\cos {135^\circ } = \cos ({180^\circ } - {45^\circ }) = \\\\ - \cos {45^\circ } = - \frac{{\sqrt 2 }}{2}\end{array}\)

الف

\(\begin{array}{l}\left\{ \begin{array}{l}0 \le \alpha \le {90^ \circ }\quad \left( i \right)\\{90^ \circ } \le \beta \le {180^ \circ }\quad \left( {ii} \right)\end{array} \right.\\\\\sin \alpha \mathop = \limits^{\left( i \right)} \sqrt {1 - {{\cos }^2}\alpha } = \sqrt {1 - {{\left( {\frac{4}{5}} \right)}^2}} = \sqrt {\frac{9}{{25}}} = \frac{3}{5}\\\\\sin \beta \mathop = \limits^{\left( {ii} \right)} \sqrt {1 - {{\cos }^2}\beta } = \sqrt {1 - {{\left( { - \;\frac{{12}}{{13}}} \right)}^2}} = \sqrt {\frac{{25}}{{169}}} = \frac{5}{{13}}\\\\\sin \left( {\alpha + \beta } \right) = \sin \alpha \;\cos \beta + \cos \alpha \;\sin \beta \\\\ = \frac{3}{5} \times \left( { - \;\frac{{12}}{{13}}} \right) + \frac{4}{5} \times \frac{5}{{13}} = \frac{{ - \;36 + 20}}{{65}} = - \;\frac{{16}}{{65}}\\\\\cos \left( {\alpha - \beta } \right) = \cos \alpha \;cos\beta + sin\alpha \;sin\beta \\\\ = \frac{4}{5} \times \left( { - \;\frac{{12}}{{13}}} \right) + \frac{3}{5} \times \frac{5}{{13}} = \frac{{ - \;48 + 15}}{{65}} = - \;\frac{{33}}{{65}}\end{array}\)

ب

\(\begin{array}{l}\left\{ \begin{array}{l}0 \le \alpha \le {90^ \circ }\\{90^ \circ } \le \beta \le {180^ \circ }\end{array} \right. \Rightarrow {90^ \circ } \le \alpha + \beta \le {270^ \circ } \Rightarrow \sin \left( {\alpha + \beta } \right) < 0\\\\ \Rightarrow {180^ \circ } < \alpha + \beta < {270^ \circ } \Rightarrow \end{array}\)

ربع سوم

الف

\(\begin{array}{l}\sin 2\alpha = 2\sin \alpha \cos \alpha \\\\\sin 2\alpha = \sin \left( {\alpha + \alpha } \right) = \sin \alpha \;\cos \alpha + \cos \;\alpha \sin \alpha \\\\ = 2\sin \alpha \;\cos \alpha \end{array}\)

ب

\(\begin{array}{l}\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha \\\\\cos 2\alpha = \cos \left( {\alpha + \alpha } \right) = \cos \alpha \;\cos \alpha - \sin \;\alpha \sin \alpha \\\\ = {\cos ^2}\alpha - {\sin ^2}\alpha \end{array}\)

پ

\(\begin{array}{l}\cos 2\alpha = 2{\cos ^2}\alpha - 1\\\\\cos 2\alpha = \cos (\alpha + \alpha ) = \\\\{\cos ^2}\alpha - {\sin ^2}\alpha = \\\\{\cos ^2}\alpha - (1 - {\cos ^2}\alpha ) = \\\\2{\cos ^2}\alpha - 1\end{array}\)

ت

\(\begin{array}{l}\cos 2\alpha = 1 - 2{\sin ^2}\alpha \\\\\cos 2\alpha = \cos (\alpha + \alpha ) = \\\\{\cos ^2}\alpha - {\sin ^2}\alpha = \\\\(1 - {\sin ^2}\alpha ) - {\sin ^2}\alpha = \\\\1 - 2{\sin ^2}\alpha \end{array}\)

نسبت مثلثاتی سینوس این زاویه به صورت زیر می شود:

\(\begin{array}{l}\alpha = 22/{5^ \circ }\\\\\cos 2\alpha = 1 - 2{\sin ^2}\alpha \\\\ \Rightarrow {\sin ^2}\alpha = \frac{{1 - \cos 2\alpha }}{2}\\\\\mathop \Rightarrow \limits^{0 < \alpha < \frac{\pi }{2}} \sin \alpha = \sqrt {\frac{{1 - \cos 2\alpha }}{2}} \\\\ \Rightarrow \sin 22/{5^ \circ } = \sqrt {\frac{{1 - \cos {{45}^ \circ }}}{2}} = \\\\\sqrt {\frac{{1 - \frac{{\sqrt 2 }}{2}}}{2}} = \frac{{\sqrt {2 - \sqrt 2 } }}{2}\end{array}\)

همچنین نسبت مثلثاتی کسینوس این زاویه به صورت زیر می شود:

\(\begin{array}{l}\alpha = 22/{5^ \circ }\\\\\cos 2\alpha = 2{\cos ^2}\alpha - 1\\\\ \Rightarrow {\cos ^2}\alpha = \frac{{1 + \cos 2\alpha }}{2}\\\\\mathop \Rightarrow \limits^{0 < \alpha < \frac{\pi }{2}} \cos \alpha = \sqrt {\frac{{1 + \cos 2\alpha }}{2}} \\\\ \Rightarrow \cos 22/{5^ \circ } = \sqrt {\frac{{1 + \cos {{45}^ \circ }}}{2}} = \\\\\sqrt {\frac{{1 + \frac{{\sqrt 2 }}{2}}}{2}} = \frac{{\sqrt {2 + \sqrt 2 } }}{2}\end{array}\)