جواب تمرین صفحه 87 درس 4 ریاضی یازدهم تجربی (مثلثات)

الف

\(\begin{array}{l}tan135^\circ {\rm{ }} + {\rm{ }}cot120^\circ {\rm{ }} = - 1 + \frac{{\sqrt 3 }}{3} = \frac{{\sqrt 3 - 3}}{3}\\\\\tan {135^ \circ } = \tan \left( {{{180}^ \circ } - {{45}^ \circ }} \right) = - \tan {45^ \circ } = - 1\\\\\cot {120^ \circ } = \cot \left( {{{180}^ \circ } - {{60}^ \circ }} \right) = - \cot {60^ \circ } = - \frac{{\sqrt 3 }}{3}\end{array}\)

ب

\(\begin{array}{l}cos\left( { - 210^\circ } \right){\rm{ }} + {\rm{ }}cot\left( {240^\circ } \right){\rm{ }} = - \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{3} = - \frac{{\sqrt 3 }}{6}\\\\\cos \left( { - {{210}^ \circ }} \right) = \cos {210^ \circ } = \cos \left( {{{180}^ \circ } + {{30}^ \circ }} \right) = - \cos {30^ \circ } = - \frac{{\sqrt 3 }}{2}\\\\\cot {240^ \circ } = \cot \left( {{{180}^ \circ } + {{60}^ \circ }} \right) = \cot {60^ \circ } = \frac{{\sqrt 3 }}{3}\end{array}\)

پ

\(\begin{array}{l}sin630^\circ {\rm{ }} + {\rm{ }}tan\left( { - 540^\circ } \right){\rm{ }} = - 1 + 0 = - 1\\\\\sin {630^ \circ } = \sin \left( {2 \times {{360}^ \circ } - {{90}^ \circ }} \right) = \sin \left( { - {{90}^ \circ }} \right) = - \sin {90^ \circ } = - 1\\\\\tan \left( { - {{540}^ \circ }} \right) = - \tan {540^ \circ } = - \tan \left( {{{360}^ \circ } + {{180}^ \circ }} \right) = \\\\ - \tan {180^ \circ } = - \tan \left( {{{180}^ \circ } - {0^ \circ }} \right) = \tan {0^ \circ } = 0\end{array}\)

ت

\(\begin{array}{l}cos\left( { - 720^\circ } \right){\rm{ }} + {\rm{ }}cot\left( { - 600^\circ } \right){\rm{ }} + {\rm{ }}tan720^\circ {\rm{ }} - {\rm{ }}tan\left( { - 600^\circ } \right){\rm{ }} = \\\\1 - \frac{{\sqrt 3 }}{3} + 0 - \sqrt 3 = \frac{{3 - 4\sqrt 3 }}{3}\\\\\cos \left( { - {{720}^ \circ }} \right) = \cos {720^ \circ } = \cos \left( {2 \times {{360}^ \circ } + {0^ \circ }} \right) = \cos {0^ \circ } = 1\\\\\cot \left( { - {{600}^ \circ }} \right) = - \cot {600^ \circ } = - \cot \left( {2 \times {{360}^ \circ } - {{120}^ \circ }} \right) = \\\\ - \cot \left( { - {{120}^ \circ }} \right) = \cot {120^ \circ } = - \frac{{\sqrt 3 }}{3}\\\\\tan {720^ \circ } = \tan \left( {2 \times {{360}^ \circ } + 0} \right) = \tan {0^ \circ } = 0\\\\\tan \left( { - {{600}^ \circ }} \right) = - \tan {600^ \circ } = \\\\ - \tan \left( {2 \times {{360}^ \circ } - {{120}^ \circ }} \right) = - \tan \left( { - {{120}^ \circ }} \right) = \tan {120^ \circ } = - \sqrt 3 \end{array}\)

ث

\(\begin{array}{l}\sin \left( {\frac{{25\pi }}{3}} \right) - \cos \left( {\frac{{23\pi }}{4}} \right) = \frac{{\sqrt 3 }}{2} - \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 3 - \sqrt 2 }}{2}\\\\\sin \left( {\frac{{25\pi }}{3}} \right) = \sin \left( {8\pi + \frac{\pi }{3}} \right)\\\\ = \sin \left( {4 \times 2\pi + \frac{\pi }{3}} \right) = \sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\\\\\cos \left( {\frac{{23\pi }}{4}} \right) = \cos \left( {6\pi - \frac{\pi }{4}} \right) = \cos \left( {3 \times 2\pi - \frac{\pi }{4}} \right)\\\\ = \cos \left( { - \frac{\pi }{4}} \right) = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}\end{array}\)

ج

\(\begin{array}{l}\frac{{\sin \frac{{3\pi }}{4} - \cos \frac{{5\pi }}{6}}}{{\sin \left( {\frac{{ - 3\pi }}{4}} \right) + \tan \left( {\frac{{ - 4\pi }}{3}} \right)}} = \frac{{\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2}}}{{ - \frac{{\sqrt 2 }}{2} - \sqrt 3 }} = - \frac{{\sqrt 2 + \sqrt 3 }}{{\sqrt 2 + 2\sqrt 3 }}\\\\\sin \frac{{3\pi }}{4} = \sin \left( {\pi - \frac{\pi }{4}} \right) = \sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}\\\\\cos \frac{{5\pi }}{6} = \cos \left( {\pi - \frac{\pi }{6}} \right) = - \cos \frac{\pi }{6} = - \frac{{\sqrt 3 }}{2}\\\\\sin \left( { - \frac{{3\pi }}{4}} \right) = - \sin \frac{{3\pi }}{4} = - \frac{{\sqrt 2 }}{2}\\\\\tan \left( {\frac{{ - 4\pi }}{3}} \right) = - \tan \frac{{4\pi }}{3} = - \tan \left( {\pi + \frac{\pi }{3}} \right)\\\\ = - \tan \frac{\pi }{3} = - \sqrt 3 \end{array}\)

الف

درست

\(\begin{array}{l}\sin {840^ \circ } = \sin \left( {{{720}^ \circ } + {{120}^ \circ }} \right) = \sin {120^ \circ }\\\\ = \sin \left( {{{180}^ \circ } - {{60}^ \circ }} \right) = \sin {60^ \circ } = \frac{{\sqrt 3 }}{2}\\\\\sin {60^ \circ } = \frac{{\sqrt 3 }}{2}\end{array}\)

ب

درست

\(\begin{array}{l}\cos \left( { - {{324}^ \circ }} \right) = \cos {324^ \circ }\\\\ = \cos \left( {{{360}^ \circ } - {{36}^ \circ }} \right) = \cos {36^ \circ }\end{array}\)

پ

درست

\(\begin{array}{l}\tan \left( { - {{1000}^ \circ }} \right) = - \tan {1000^ \circ }\\\\ = - \tan \left( {{{1080}^ \circ } - {{80}^ \circ }} \right) = \\\\ - \tan \left( {3 \times {{360}^ \circ } - {{80}^ \circ }} \right)\\\\ = - \tan \left( { - {{80}^ \circ }} \right) = \tan {80^ \circ }\end{array}\)

ت

درست

\(\sin {875^ \circ } = \sin \left( {{{720}^ \circ } + {{155}^ \circ }} \right) = \sin {155^ \circ }\)

با توجه به رابطه نوشته شده، باید زوایا متمم هم باشند؛ بنابراین:

\(x + \left( {{{20}^ \circ } + x} \right) = {90^ \circ } \Rightarrow 2x = {70^ \circ } \Rightarrow x = {35^ \circ }\)

با توجه به رابطه نوشته شده، باید زوایا متمم هم باشند؛ بنابراین:

\(\begin{array}{l}\left( {x + \frac{\pi }{{18}}} \right) + \left( {\frac{{2\pi }}{9} + x} \right) = \frac{\pi }{2} \Rightarrow 2x + \frac{{5\pi }}{{18}} = \frac{\pi }{2}\\\\ \Rightarrow 2x = \frac{{4\pi }}{{18}} \Rightarrow x = \frac{\pi }{9}\end{array}\)

می توان زوایای بسیاری یافت که در تساوی های فوق صدق کنند؛ کافی است ضرایب π2 رادیان و یا 360 درجه را به این زوایا اضافه کرد؛ مثلا برای قسمت اول، 395درجه و برای قسمت دوم، \((2\pi + \frac{\pi }{9} = )\frac{{19\pi }}{9}\) رادیان جواب های دیگری هستند.