Bài 1 trang 153 sgk đại số 10. Tính a) \(\cos {225^0},\sin {240^0},cot( - {15^0}),tan{75^0}\); b) \(\sin \frac{7\pi}{12}\), \(\cos \left ( -\frac{\pi}{12} \right )\), \(\tan\left ( \frac{13\pi}{12} \right )\) Giải a) +) \(\cos{225^0} = \cos({180^0} +{45^0})= - \cos{45^{0}}= -\frac{\sqrt{2}}{2}\) +) \(\sin{240^0} = \sin({180^0} +{60^0}) = - \sin{60^0}= -\frac{\sqrt{3}}{2}\) +) \(\cot( - {15^0})= - \cot{15^0} = - \tan{75^0} =- \tan({30^0} +{45^0})\) \( =\frac{-\tan30^{0}-\tan45^{0}}{1-\tan30^{0}\tan45^{0}}=\frac{-\frac{1}{\sqrt{3}}-1}{1-\frac{1}{\sqrt{3}}}=-\frac{\sqrt{3}+1}{\sqrt{3}-1}=-\frac{(\sqrt{3}+1)^{2}}{2} = -2 - \sqrt 3\) +) \(\tan 75^0= \cot15^0= 2 + \sqrt3\) b) +) \(\sin \frac{7\pi}{12} = \sin \left ( \frac{\pi}{3}+\frac{\pi}{4} \right ) =\sin\frac{\pi }{3}\cos\frac{\pi}{4}+ \cos \frac{\pi }{3}\sin\frac{\pi}{4}\) \( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )=\frac{\sqrt{6}+\sqrt{2}}{4}\) +) \(\cos \left ( -\frac{\pi }{12} \right ) = \cos \left ( \frac{\pi }{4} -\frac{\pi }{3}\right ) = \cos \frac{\pi }{4}\cos\frac{\pi }{3} + sin \frac{\pi }{3}sin \frac{\pi }{4}\) \( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )=0,9659\) +) \(\tan \left ( \frac{13\pi }{12} \right ) = \tan(π + \frac{\pi }{12}) = \tan \frac{\pi }{12} = \tan \left ( \frac{\pi }{3}-\frac{\pi}{4} \right )\) \(= \frac{\tan\frac{\pi }{3}-\tan\frac{\pi }{4}}{1+\tan\frac{\pi }{3}\tan\frac{\pi }{4}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}= 2 - \sqrt3\) Bài 2 trang 154 sgk đại số 10. Tính a) \(\cos(α + \frac{\pi}{3}\)), biết \(\sinα = \frac{1}{\sqrt{3}}\) và \(0 < α < \frac{\pi }{2}\). b) \(\tan(α - \frac{\pi }{4}\)), biết \(\cosα = -\frac{1}{3}\) và \( \frac{\pi }{2} < α < π\) c) \(\cos(a + b), \sin(a - b)\) biết \(\sin a = \frac{4}{5}\), \(0^0< a < 90^0\) và \(\sin b = \frac{2}{3}\), \(90^0< b < 180^0\) Giải a) Do \(0 < α < \frac{\pi}{2}\) nên \(\sinα > 0, \cosα > 0\) \(\cosα = \sqrt{1-\sin^{2}\alpha }=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\) \(cos(α + \frac{\pi}{3}) = \cosα\cos \frac{\pi }{3} - \sinα\sin \frac{\pi}{3}\) \( = \frac{\sqrt{6}}{3}.\frac{1}{2}-\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{2}=\frac{\sqrt{6}-3}{6}\) b) Do \( \frac{\pi}{2}< α < π\) nên \(\sinα > 0, \cosα < 0, \tanα < 0, \cotα < 0\) \(\tanα = -\sqrt{\frac{1}{cos^{2}\alpha }-1}=-\sqrt{3^{3}-1} = -2\sqrt2\) \(tan(α - \frac{\pi}{4}) = \frac{\tan\alpha -\tan\frac{\pi}{4}}{1+\tan\alpha tan\frac{\pi}{4}}=\frac{-1-2\sqrt{2}}{1-2\sqrt{2}}=\frac{2\sqrt{2}+1}{2\sqrt4{2}-1}\) c) \(0^0< a < 90^0\Rightarrow \sin a > 0, \cos a > 0\) \(90^0< b < 180^0\Rightarrow \sin b > 0, \cos b < 0\) \(\cos a = \sqrt{1-sin^{2}a}=\sqrt{1-\left ( \frac{4}{5} \right )^{2}}=\frac{3}{5}\) \(\cos b = -\sqrt{1-sin^{2}a}=-\sqrt{1-\left ( \frac{2}{3} \right )^{2}}=-\frac{\sqrt{5}}{3}\) \(\cos(a + b) = \cos a\cos b - \sin a\sin b\) \( =\frac{3}{5}\left ( -\frac{\sqrt{5}}{3} \right )-\frac{4}{5}.\frac{2}{3}=-\frac{3\sqrt{5}+8}{15}\) \(\eqalign{ & \sin(a - b) = \sin a\cos b - \cos a\sin b \cr & = {4 \over 5}.\left( { - {{\sqrt 5 } \over 3}} \right) - {3 \over 5}.{2 \over 3} = - {{4\sqrt 5 + 6} \over {15}} \cr} \) Bài 3 trang 154 sgk đại số 10. Rút gọn các biểu thức a) \(\sin(a + b) + \sin(\frac{\pi}{2}- a)\sin(-b)\). b) \(cos(\frac{\pi }{4} + a)\cos( \frac{\pi}{4} - a) + \frac{1 }{2} \sin^2a\) c) \(\cos( \frac{\pi}{2} - a)\sin( \frac{\pi}{2} - b) - \sin(a - b)\) Giải a) \(\sin(a + b) + \sin( \frac{\pi }{2} - a)\sin(-b) = \sin a\cos b + \cos a\sin b - \cos a\sin b = \sin a\cos b\) b) \(\cos( \frac{\pi }{4} + a)\cos(\frac{\pi }{4}- a) + \frac{1 }{2}\sin^2a\) \( =\frac{1 }{2}\cos\left [ \frac{\pi }{4}+a+\frac{\pi}{4} -a\right ]+\frac{1}{2}\cos\left [ \left ( \frac{\pi }{4} +a\right ) -\left ( \frac{\pi}{4}-a \right )\right ]+\frac{1}{2}\left ( \frac{1-\cos 2a}{2} \right )\) \( =\frac{1}{2}\cos 2a + \frac{1}{4}(1 - \cos 2a) = \frac{1+\cos 2a}{4 }= \frac{1 }{2}\cos^2 a\) c) \(\cos( \frac{\pi}{2} - a)\sin( \frac{\pi}{2} - b) - \sin(a - b) = \sin a\cos b - \sin a\cos b + \sin b\cos a\) \(= \sin b\cos a\) Bài 4 trang 154 sgk đại số 10. Chứng minh các đẳng thức a) \( \frac{cos(a-b)}{cos(a+b)}=\frac{cotacotb+1}{cotacotb-1}\) b) \(\sin(a + b)\sin(a - b) = \sin^2a – \sin^2b = \cos^2b – \cos^2a\) c) \(\cos(a + b)\cos(a - b) = \cos^2a - \sin^2b = \cos^2b – \sin^2a\) Giải a) \(VT = {{\cos a\cos b+\sin a\sin b}\over{\cos a\cos b-\sin a\sin b}}=\frac{\frac{\cos a\cos b}{\sin a\sin b}+1}{\frac{\cos a\cos b}{\sin a\sin b}-1}=\frac{\cot a\cot b+1}{\cot a\cot b-1}\) b) \(VT = [\sin a\cos b + \cos a\sin b][\sin a\cos b - \cos a\sin a]\) \(= (\sin a\cos b)^2– (\cos a\sin b)^2= \sin^2 a(1 – \sin^2 b) – (1 – \sin^2 a)\sin^2 b\) \(= \sin^2a – \sin^2b = \cos^2b( 1– \cos^2a) – \cos^2 a(1 – \cos^2 b) = \cos^2 b – \cos^2 a\) c) \(VT = (\cos a\cos b - \sin a\sin b)(\cos a\cos b + \sin a\sin b)\) \(= (\cos a\cos b)^2 – (\sin a\sin b)^2\) \(= \cos^2 a(1 – \sin^2 b) – (1 – \cos^2 a)\sin^2 b = \cos^2 a – \sin^2 b\) \(= \cos^2 b(1 – \sin^2 a) – (1 – \cos^2 b)\sin^2 a = \cos^2 b – \sin^2 a\) Bài 5 trang 154 sách giáo khoa Đại Số 10. Tính \(\sin2a, \cos2a, \tan2a\), biết a) \(\sin a = -0,6\) và \(π < a < {{3\pi } \over 2}\) b) \(\cos a = - {5 \over {13}}\) và \({\pi \over 2} < a < π\) c) sina + cosa = \({1 \over 2}\) và \({{3\pi } \over 4}\) < a < π Giải a) \(\sin a = -0,6\) và \(\pi < a < {{3\pi } \over 2}\) \(\sin 2a = 2\sin a\cos a\) (1) (công thức) Mà \(\pi < a < {{3\pi } \over 2} \Rightarrow \cos a < 0\) và \(\sin a = -0,6 \Rightarrow \cos a = - {4 \over 5}\) \((1) \Leftrightarrow \sin 2{\rm{a}} = 2.( - 0,6).\left( { - {4 \over 5}} \right) \Leftrightarrow \sin 2{\rm{a}} = {{24} \over {25}}\) \(\cos 2a = 1 - 2\sin^2a = 1 - 2{\left( { - {3 \over 5}} \right)^2} = 1 - {{18} \over {25}}\) \(\cos 2a = {7 \over {25}}\) \(\tan 2a = {{\sin 2a} \over {\cos 2a}} = {{24} \over {25}}.{{25} \over 7} = {{24} \over 7}\) b) \(\cos a = - {5 \over {13}}\) và \({\pi \over 2} < a < \pi\) Vì \({\pi \over 2} < a < \pi\) nên \(\sin a > 0; \tan a < 0\) và \(\cos a = - {5 \over {13}}\) nên \(\sin {\rm{a}} = {{12} \over {13}}\) Do đó, \(\sin 2{\rm{a}} = 2.{{12} \over {13}}.\left( { - {5 \over {13}}} \right) = - {{120} \over {169}}\) \(\cos 2a = 2.{\cos ^2}a - 1 = 2.{{25} \over {169}} - 1 = - {{119} \over {169}}\) \(\tan 2a = {{\sin 2a} \over {\cos 2a}} = \left( { - {{120} \over {169}}} \right).\left( { - {{169} \over {119}}} \right) = {{120} \over {119}}\) c) \(\sin {\rm{a}} + {\mathop{\rm cosa}\nolimits} = {1 \over 2}\) và \({{3\pi } \over 4} < a < \pi\) Vì \({{3\pi } \over 4} < a < \pi \) nên \(\sin a > 0; \cos a < 0\) \(\left. \matrix{{\cos ^2}a + {\sin ^2}a = 1 \hfill \cr \sin a + \cos a = {1 \over 2} \hfill \cr} \right\} \Rightarrow \left\{ \matrix{\cos a = {{1 - \sqrt 7 } \over 4} \hfill \cr \sin a = {{1 + \sqrt 7 } \over 4} \hfill \cr} \right.\) Suy ra : \(\sin 2a = 2.{{1 + \sqrt 7 } \over 4}.{{1 - \sqrt 7 } \over 4} = {{ - 3} \over 4}\) \(\cos 2a = 1 - 2{\sin ^2}a = 1 - 2{\left( {{{1 + \sqrt 7 } \over 4}} \right)^2} = {{ \sqrt 7 } \over 4}\) \(\tan 2a = - {{3\sqrt 7 } \over 7}\) Bài 6 trang 154 sgk đại số 10. Cho \(\sin 2a = - {5 \over 9}\) và \({\pi \over 2}< a < π\). Tính \(\sin a\) và \(\cos a\). Giải \({\pi \over 2}< a < π\Rightarrow \sin a > 0, \cos a < 0\) \(\cos 2a = \pm \sqrt {1 - {{\sin }^2}2a} = \pm \sqrt {1 - {{\left( {{5 \over 9}} \right)}^2}} = \pm {{2\sqrt {14} } \over 9}\) Nếu \(\cos 2a = {{2\sqrt {14} } \over 9}\) thì \(\eqalign{ & \sin a = \sqrt {{{1 - \cos 2a} \over 2}} = \sqrt {{{1 - {{2\sqrt {14} } \over 9}} \over 2}} = {{\sqrt {9 - 2\sqrt {14} } } \over {3\sqrt 2 }} \cr & = {{\sqrt {{{\left( {\sqrt 7 - \sqrt 2 } \right)}^2}} } \over {3\sqrt 2 }} = {{\sqrt 7 - \sqrt 2 } \over {3\sqrt 2 }} = {{\sqrt {14} - 2} \over 6} \cr} \) \(\cos a = - \sqrt {{{1 + \cos 2a} \over 2}} = - {{\sqrt {14} + 2} \over 6}\) Nếu \(\cos 2a = -{{2\sqrt {14} } \over 9}\) thì \(\eqalign{ & \sin a = \sqrt {{{1 - \cos 2a} \over 2}} = \sqrt {{{1 + {{2\sqrt {14} } \over 9}} \over 2}} = {{\sqrt {9 + 2\sqrt {14} } } \over {3\sqrt 2 }} \cr & = {{\sqrt {{{\left( {\sqrt 7 + \sqrt 2 } \right)}^2}} } \over {3\sqrt 2 }} = {{\sqrt 7 + \sqrt 2 } \over {3\sqrt 2 }} = {{\sqrt {14} + 2} \over 6} \cr & \cos a = - \sqrt {{{1 + \cos 2a} \over 2}} = {{ - \sqrt {14} + 2} \over 6} \cr} \) Bài 7 trang 155 sgk đại số 10. Biến đổi thành tích các biểu thức sau a) \(1 - \sin x\); b) \(1 + \sin x\); c) \(1 + 2\cos x\); d) \(1 - 2\sin x\) Giải a) \(1 - \sin x = \sin \frac{\pi }{2} - \sin x = 2\cos \frac{\frac{\pi }{2}+x}{2}\sin \frac{\frac{\pi}{2}-x}{2}\) \(= 2 \cos \left ( \frac{\pi }{4} +\frac{x}{2}\right )\sin\left ( \frac{\pi }{4} -\frac{x}{2}\right )\) b) \(1 + \sin x = \sin \frac{\pi }{2} + \sin x = 2\sin \left ( \frac{\pi }{4} +\frac{x}{2}\right )\cos \left ( \frac{\pi }{4} -\frac{x}{2}\right )\) c) \(1 + 2\cos x = 2( \frac{1}{2} + \cos x) = 2(\cos \frac{\pi}{3} + \cos x) \) \(= 4\cos \left ( \frac{\pi }{6} +\frac{x}{2}\right )\cos \left ( \frac{\pi }{6} -\frac{x}{2}\right )\) d) \(1 - 2\sin x = 2( \frac{1}{2} - \sin x) = 2(\sin \frac{\pi}{6} - \sin x)\) \(= 4\cos \left ( \frac{\pi }{12} +\frac{x}{2}\right )\sin \left ( \frac{\pi }{12} -\frac{x}{2}\right )\) Bài 8 trang 155 sgk đại số 10. Rút gọn biểu thức \(A = {{{\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 3{\rm{x}} + \sin 5{\rm{x}}} \over {{\mathop{\rm cosx}\nolimits} + cos3x + cos5x}}\). Lời giải: Ta có: +) \(\sin x + \sin 3x + \sin 5x = \sin x + \sin 5x + \sin 3x\) \(= 2\sin {{x + 5x} \over 2}.\cos {{x - 5x} \over 2} + \sin 3x = 2\sin 3x + \cos 2x + \sin 3x\) \(= \sin 3x (2\cos 2x + 1)\) (1) +) \( \cos x + \cos3x + \cos5x = \cos x + \cos5x +\cos3x = 2\cos3x . \cos2x + \cos3x = \cos3x (2\cos2x + 1)\) (2) Từ (1) và (2) ta có: \(A = {{\sin 3x} \over {\cos 3x}} = \tan 3x\) Vậy \(A= \tan 3x\)
