نمونه سوالات فصل 4 حسابان یازدهم همراه با جواب تشریحی

\(a = \frac{1}{r} \Rightarrow a = \frac{8}{{10}} \Rightarrow a = 0/8\)

\(\frac{\pi }{{20}} \Rightarrow \frac{{180}}{{20}} = 9\)

\(\begin{array}{l}L = 1200km\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r = 6400km\,\,\,\,\,\,\,\,\,\,\,\theta = ?\\\\L = r\theta \Rightarrow 1200 = 6400\theta \Rightarrow \theta = \frac{{12}}{{64}} = \frac{3}{{16}}\end{array}\)

\(\theta = 120 \Rightarrow \theta = \frac{{2\pi }}{3} \Rightarrow \theta = \frac{L}{r} \Rightarrow \frac{{2\pi }}{3} = \frac{L}{{18}} \Rightarrow L = 12\pi \)

\(\theta = 150 \Rightarrow \theta = \frac{{5\pi }}{6} \Rightarrow \theta = \frac{L}{r} \Rightarrow \frac{{5\pi }}{6} = \frac{L}{{24}} \Rightarrow L = 20\pi \)

\(\begin{array}{l}\frac{{40}}{\alpha } = \frac{{180}}{\pi } \Rightarrow \alpha = \frac{{40\pi }}{{180}} = \frac{{2\pi }}{9}rad\\\\\alpha = \frac{{AB}}{r} \Rightarrow \frac{{2\pi }}{9} = \frac{{AB}}{4} \Rightarrow AB = \frac{{8\pi }}{9}cm\end{array}\)

در قرقره بزرگتر داریم :

\(\theta = \frac{{P{P^`}}}{r} \Rightarrow P{P^`} = \frac{\pi }{3} \times 8 = \frac{{8\pi }}{3}cm\)

چون دو قرقره با یک تسمه به هم متصل شده اند پس قرقره کوچکتر نیز \(\frac{{8\pi }}{3}cm\) حرکت می کند بنابراین :

\(\theta = \frac{L}{r} \Rightarrow \theta = \frac{{\frac{{8\pi }}{3}}}{3} = \frac{{8\pi }}{9}rad\)

\(\theta = 210 \Rightarrow \theta = \frac{{7\pi }}{6} \Rightarrow \theta = \frac{I}{r} \Rightarrow \frac{{7\pi }}{6} = \frac{I}{{15}} \Rightarrow I = \frac{{35\pi }}{2}\)

\(\frac{{ - 3}}{\pi } = \frac{\alpha }{{180}} \Rightarrow \alpha = \frac{{ - 540}}{\pi }\)

\(\begin{array}{l}\sin 840 = \sin 60\\\\\sin 840 = \sin (2 \times 360 + 120) = \sin 120 = \sin (180 - 60)\\\\ = \sin 60\end{array}\)

\(\begin{array}{l}\cos ( - 324) = \cos 36\\\\\cos ( - 324)\cos 324 = \cos (360 - 36) = \cos 36\end{array}\)

\(\begin{array}{l}tg( - 1000) = tg80\\\\tg( - 1000) = - tg1000 = - tg(3 \times 360 - 80)\\\\ = - ( - tg80) = tg80\end{array}\)

\(\begin{array}{l}\sin 875 = \sin 155\\\\\sin 875 = \sin (2 \times 360 + 155) = \sin 155\end{array}\)

\(\begin{array}{l}tg135 + \cot g120 = tg(180 - 45) + \cot g(180 - 60)\\\\ = - tg45 + ( - \cot g60) = - 1 - \frac{{\sqrt 3 }}{3}\end{array}\)

\(\begin{array}{l}\cos ( - 210) + \cot g(240) = \cos 210 + \cot g(240)\\\\ = \cos (180 + 30) + \cot g(180 + 60) = - \cos 30 + \cot g60\\\\ = - \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{3}\end{array}\)

\(\begin{array}{l}\sin 630 + tg(540) = \sin 630 - tg540\\\\ = \sin (2 \times 360 - 90) - tg(360 + 180) = - \sin 90 - tg180\\\\ = - 1 + 0 = - 1\end{array}\)

\(\begin{array}{l}\cos ( - 720) + \cot g( - 600) + tg720 - tg( - 600) = \\\\ = \cos 720 - \cot g600 + tg720 + tg600\\\\\cos (2 \times 360 + 0) - \cot g(2 \times 360 - 120) + tg(2 \times 360 + 0)\\\\ + tg(2 \times 360 - 120) = \cos 0 - ( - \cot g120) + tg0 - tg120\\\\\cos 0 + \cot g(180 - 60) + tg0 - tg(180 - 60)\\\\ = \cos 0 - \cot g60 + tg0 + tg60 = 1 - \frac{{\sqrt 3 }}{3} + 0 + \sqrt 3 \\\\ = 1 - \frac{{\sqrt 3 }}{3} + \sqrt 3 \end{array}\)

\(\begin{array}{l}\sin (\frac{{25\pi }}{3}) - \cos (\frac{{23\pi }}{4}) = \sin (8\pi + \frac{\pi }{3}) - \cos (8\pi - \frac{\pi }{3})\\\\ = \sin \frac{\pi }{3} - \cos \frac{\pi }{3} = \frac{{\sqrt 3 }}{2} - \frac{1}{2}\end{array}\)

\(\begin{array}{l}\frac{{\sin \frac{{3\pi }}{4} - \cos \frac{{5\pi }}{6}}}{{\sin (\frac{{ - 3\pi }}{4}) + tg(\frac{{ - 4\pi }}{3})}} = \frac{{\sin (\pi - \frac{\pi }{4}) - \cos (\pi - \frac{\pi }{6})}}{{ - \sin (\pi - \frac{\pi }{4}) - tg(\pi + \frac{\pi }{3})}}\\\\ = \frac{{\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2}}}{{ - \frac{{\sqrt 2 }}{2} - \sqrt 3 }} = \frac{{ - 4 - \sqrt 6 }}{{10}}\end{array}\)

\(\cos (315) - \sin (\frac{{5\pi }}{4}) = \cos 45 + \sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \sqrt 2 \)

\(\begin{array}{l}\frac{{\sin (180 - 20) - \cos (180 + 20)}}{{\cos (90 + 20) + \sin (90 - 20)}} = \frac{{\sin 20 + \cos 20}}{{ - \sin 20 + \cos 20}}\\\\ \Rightarrow \frac{{tg20 + 1}}{{ - tg20 + 1}} = \frac{{1/36}}{{0/64}} = \frac{{17}}{8}\end{array}\)

\(\sin \frac{{5\pi }}{4} = \sin (\pi + \frac{\pi }{4}) = - \sin \frac{\pi }{4} = - \frac{{\sqrt 2 }}{2}\)

\(\cos \frac{{9\pi }}{4} = \cos (2\pi + \frac{\pi }{4}) = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}\)

\(\begin{array}{l}\cot g(750) = \cot g(720 + 30) = \cot g(4\pi + \frac{\pi }{6})\\\\ = \cot g\frac{\pi }{6} = \sqrt 3 \end{array}\)

\(tg( - 150) = - tg150 = - t(\pi - \frac{\pi }{6}) = tg\frac{\pi }{6} = \frac{{\sqrt 3 }}{3}\)

\(\begin{array}{l}\frac{{tg(\frac{{7\pi }}{4}) - tg(\frac{{5\pi }}{4})}}{{1 + tg(\frac{\pi }{3}) \times \cot g(\frac{{4\pi }}{3})}} = \frac{{tg(2\pi - \frac{\pi }{4}) - tg(\pi + \frac{\pi }{4})}}{{1 + tg(\frac{\pi }{3}) \times \cot g(\pi + \frac{\pi }{3})}}\\\\ = \frac{{ - tg(\frac{\pi }{4}) - tg(\frac{\pi }{4})}}{{1 + tg(\frac{\pi }{3}) \times \cot g(\frac{\pi }{3})}} = \frac{{ - 1 - 1}}{{1 + \sqrt 3 \times \frac{1}{{\sqrt 3 }}}} = \frac{{ - 2}}{2} = - 1\end{array}\)

\(\sin (\frac{{27\pi }}{2} + \theta ) = - \cos \theta \)

\(\cos ( - \frac{\pi }{2} - \theta ) = - \sin \theta \)

\(\cot g(93\pi - \theta ) = - \cot g\theta \)

\(\begin{array}{l}\sin ( - \frac{{41\pi }}{4}) = \sin (\frac{{ - 40\pi - \pi }}{4}) = \sin ( - 10\pi - \frac{\pi }{4}) = - \sin \frac{\pi }{4}\\\\ = - \frac{{\sqrt 2 }}{2}\end{array}\)

\(\cos (\frac{{13\pi }}{6})( = \cos (\frac{{12\pi + \pi }}{6}) = \cos (2\pi + \frac{\pi }{6}) = \cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}\)

\(tg(\frac{{29\pi }}{3}) = tg(\frac{{30\pi - \pi }}{3}) = tg(10\pi - \frac{\pi }{3}) = - tg\frac{\pi }{3} = - \sqrt 3 \)

\(\begin{array}{l}\cot g( - \frac{{14\pi }}{3}) = \cot g(\frac{{ - 15\pi + \pi }}{3}) = \cot g( - 5\pi + \frac{\pi }{3})\\\\ = - \cot g\frac{\pi }{3} = - \frac{{\sqrt 3 }}{3}\end{array}\)

\(\begin{array}{l}f(\frac{{5\pi }}{3}) = 2\sin (\frac{{5\pi }}{3} + \frac{\pi }{6}) = 2\sin (\frac{{10\pi + \pi }}{6}) = 2\sin (\frac{{11\pi }}{6})\\\\ = 2\sin (\frac{{12\pi - \pi }}{6}) = 2\sin (2\pi - \frac{\pi }{6}) = - 2\sin (\frac{\pi }{6}) = - 2 \times \frac{1}{2} = - 1\end{array}\)

\(\begin{array}{l}f(\frac{{19\pi }}{6}) = 2\sin (\frac{{19\pi }}{6} + \frac{\pi }{6}) = 2\sin (\frac{{20\pi }}{6}) = 2\sin (\frac{{10\pi }}{3})\\\\ = \sin (\frac{{9\pi + \pi }}{3}) = 2\sin (3\pi + \frac{\pi }{3}) = - 2\sin (\frac{\pi }{3})\\\\ = - 2 \times \frac{{\sqrt 3 }}{2} = - \sqrt 3 \end{array}\)