جواب تمرین صفحه 48 درس 2 ریاضی دوازدهم تجربی (مثلثات)

الف 

\(\begin{array}{l}{\sin ^2}\alpha = 1 - {(\frac{5}{{13}})^2} = \frac{{144}}{{169}} \Rightarrow \sin \alpha = \frac{{12}}{{13}}\\\\\cos 2\alpha = 2{\cos ^2}\alpha - 1 = 2 \times \frac{{25}}{{169}} - 1 = - \frac{{119}}{{169}}\end{array}\)

ب 

\(\sin 2\alpha = 2\sin \alpha \cos \alpha = 2 \times \frac{{12}}{{13}} \times \frac{5}{{13}} = \frac{{120}}{{169}}\)

\(\begin{array}{l}{\sin ^2}22/{5^ \circ } = \frac{{1 - \cos {{45}^ \circ }}}{2} = \frac{{1 - \frac{{\sqrt 2 }}{2}}}{2} = \frac{{2 - \sqrt 2 }}{4}\\\\ \Rightarrow \sin 22/{5^ \circ } = \frac{{\sqrt {2 - \sqrt 2 } }}{2}\\\\{\cos ^2}22/{5^ \circ } = \frac{{1 + \cos {{45}^ \circ }}}{2} = \frac{{1 + \frac{{\sqrt 2 }}{2}}}{2} = \frac{{2 + \sqrt 2 }}{4}\\\\ \Rightarrow \cos 22/{5^ \circ } = \frac{{\sqrt {2 + \sqrt 2 } }}{2}\end{array}\)

الف 

\(\begin{array}{l}\sin \frac{\pi }{2} = \sin 3x \Rightarrow 3x = 2k\pi + \frac{\pi }{2}\,\,,\;k \in \mathbb{Z}\,\\\\ \Rightarrow x = \frac{{2k\pi }}{3} + \frac{\pi }{6}\;,\;k \in \mathbb{Z}\end{array}\)

ب 

\(\begin{array}{l}cos2x_{}^{}cosx{\rm{ }} + {\rm{ }}1{\rm{ }} = {\rm{ }}0 \Rightarrow 2{\cos ^2} - 1 - \cos x + 1 = 0\\\\ \Rightarrow 2{\cos ^2} - \cos x = 0\\\\ \Rightarrow \cos x(2\cos x - 1) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}\cos x = 0\quad \Rightarrow \quad x = k\pi + \frac{\pi }{2}\\\cos x = \frac{1}{2}\quad \Rightarrow \quad x = 2k\pi \pm \frac{\pi }{3}\end{array} \right.\end{array}\)

پ 

\(\begin{array}{l}cosx{\rm{ }} = {\rm{ }}cos2x \Rightarrow x = 2k\pi \pm 2x\\\\ \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + 2x \Rightarrow x = - 2k\pi \\x = 2k\pi - 2x \Rightarrow x = \frac{{2k\pi }}{3}\end{array} \right.\end{array}\)

ت 

\(\begin{array}{l}cos2x--3sinx{\rm{ }} + {\rm{ }}1{\rm{ }} = {\rm{ }}0 \Rightarrow - 2{\sin ^2}x + 1 - 3\sin x + 1 = 0\\\\ \Rightarrow - 2{\sin ^2}x - 3\sin x + 2 = 0\\\\ \Rightarrow \Delta = {b^2} - 4ac = 25\\ \Rightarrow \sin x = \frac{{3 \pm \sqrt {25} }}{{ - 4}}\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = \frac{{3 + 5}}{{ - 4}} = - 2\;\;\; \otimes \\\sin x = \frac{{3 - 5}}{{ - 4}} = - \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}x = 2k\pi - \frac{\pi }{6}\\x = 2k\pi + \pi + \frac{\pi }{6}\end{array} \right.\end{array} \right.\end{array}\)

ث 

\(\begin{array}{l}{\cos ^2}x - \sin x = \frac{1}{4} \Rightarrow 1 - {\sin ^2}x - \sin x = \frac{1}{4}\\\\ \Rightarrow {\sin ^2}x + \sin x - \frac{3}{4} = 0 \Rightarrow \Delta = {b^2} - 4ac = 4\\\\ \Rightarrow \sin x = \frac{{ - 1 \pm \sqrt 4 }}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = \frac{{ - 1 + 2}}{2} = \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{6}\\x = 2k\pi + \pi - \frac{\pi }{6}\end{array} \right.\\\sin x = \frac{{ - 1 - 2}}{2} = - \frac{3}{2}\quad \otimes \end{array} \right.\end{array}\)

ج

\(\begin{array}{l}sinx - cos2x = 0 \Rightarrow \sin x + 2{\sin ^2}x - 1 = 0\\\\ \Rightarrow \Delta = {b^2} - 4ac = 9\\\\ \Rightarrow \sin x = \frac{{ - 1 \pm \sqrt 9 }}{4}\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = \frac{{ - 1 + 3}}{4} = \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{6}\\x = 2k\pi + \pi - \frac{\pi }{6}\end{array} \right.\\\sin x = \frac{{ - 1 - 3}}{4} = - 1 \Rightarrow x = 2k\pi + \frac{{3\pi }}{2}\end{array} \right.\end{array}\)

دو مثلث می توان ساخت :

\(\begin{array}{l}S = \frac{1}{2} \times 2 \times 6 \times \sin \alpha = 3\\\\ \Rightarrow \sin \alpha = \frac{1}{2} = \sin \frac{\pi }{6}\\\\ \Rightarrow \alpha = \frac{\pi }{6}\quad ,\quad \frac{{5\pi }}{6}\end{array}\)