Bài 9 trang 161 SGK Đại số 10

Câu hỏi: Tính

Câu a​

\(4(cos{24^0} + \cos {48^0} - \cos {84^0} - \cos {12^0})\)
Lời giải chi tiết:
$\cos 24^{0}+\cos 48^{0}$
$=\cos \left(36^{0}-12^{0}\right)+\cos \left(36^{0}+12^{0}\right)$
$=2 \cos 36^{0} \cos 12^{0}$
$\cos 84^{0}+\cos 12^{0}=2 \cos 36^{0} \cos 48^{0}$
$4\left(\cos 24^{0}+\cos 48^{0}-\cos 84^{0}-\cos 12^{0}\right)$
$=8 \cos 36^{0}\left(\cos 12^{0}-\cos 48^{0}\right)$
$=8 \cos 36^{0} \cdot 2 \sin 30^{0} \cdot \sin 18^{0}$
$=8 \cos 36^{0} \sin 18^{0}$
$=8 \cos 36^{0} \cdot \sqrt{\frac{1-\cos 36^{0}}{2}}$
Đặt \(36^0= x\) ta có:
\(\eqalign{
& sin3x{\rm{ }} = {\rm{ }}sin{\rm{ }}\left({{{180}^0} - 3x} \right) = sin2x \cr
& \Leftrightarrow 3\sin x - 4{\sin ^3}x = 2\sin x\cos x \cr
& \Leftrightarrow 3 - 4(1 - {\cos ^2}x) = 2{\mathop{\rm cosx}\nolimits} \cr
& \Leftrightarrow 4co{s^2}x - 2\cos x - 1 = 0 \cr
& \Rightarrow {\mathop{\rm cosx}\nolimits} = \cos {36^0} = {{1 + \sqrt 5 } \over 4} \cr} \)
Vậy : \(4(cos{24^0} + \cos {48^0} - \cos {84^0} - \cos {12^0}) \)\(= 2(1 + \sqrt 5)\sqrt {{{3 - \sqrt 5 } \over 8}} = 2\)

Câu b​

\(96\sqrt 3 \sin {\pi \over {48}}\cos {\pi \over {48}}\cos {\pi \over {24}}\cos {\pi \over {12}}\cos {\pi \over 6}\)
Lời giải chi tiết:
$96 \sqrt{3} \sin \frac{\pi}{48} \cos \frac{\pi}{48} \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6}$
$=48 \sqrt{3} \sin \frac{\pi}{24} \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6}$
$=24 \sqrt{3} \sin \frac{\pi}{12} \cos \frac{\pi}{12} \cos \frac{\pi}{6}$
$=12 \sqrt{3} \sin \frac{\pi}{6} \cos \frac{\pi}{6}=6 \sqrt{3} \sin \frac{\pi}{3}=9$

Câu c​

\(\tan {9^0} - \tan {63^0} + \tan {81^0} - \tan {27^0}\)
Lời giải chi tiết:
$\tan 9^{0}-\tan 63^{0}+\tan 81^{0}-\tan 27^{0}$
$=\frac{\cos 81^{0}}{\sin 81^{0}}+\frac{\sin 81^{0}}{\cos 81^{0}}-\left(\frac{\cos 27^{0}}{\sin 27^{0}}+\frac{\sin 27^{0}}{\cos 27^{0}}\right)$
$=\frac{1}{\sin 81^{0} \cdot \cos 81^{0}}-\frac{1}{\sin 27^{0} \cdot \cos 27^{0}}$
$=\frac{2}{\sin 18^{0}}-\frac{2}{\sin 54^{0}}=\frac{2}{\cos 72^{0}}-\frac{2}{\cos 36^{0}}$
$=\frac{2}{2 \cos ^{2} 36^{0}-1}-\frac{2}{\cos 36^{0}}$
Thay \(\cos {36^0} = {{1 + \sqrt 5 } \over 4}\) ta được: \(\tan {9^0} - \tan {63^0} + \tan {81^0} - \tan {27^0} \)\(= 4\)
Cách khác:
Ta có : $\tan 9^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}-\tan 27^{\circ}$
$=\tan 9^{\circ}-\cot 27^{\circ}+\cot 9^{\circ}-\tan 27^{\circ}$
$=\tan 9^{\circ}+\cot 9^{\circ}-\left(\cot 27^{\circ}+\tan 27^{\circ}\right)$
$=\tan 9^{\circ}+\frac{1}{\tan 9^{\circ}}-\left(\tan 27^{\circ}+\frac{1}{\tan 27^{\circ}}\right)$
$=\frac{\tan ^{2} 9^{\circ}+1}{\tan 9^{\circ}}-\frac{\tan ^{2} 27^{\circ}+1}{\tan 27^{\circ}}$
$=\frac{1}{\cos 9^{\circ} \cdot \sin 9^{\circ}}-\frac{1}{\cos 27^{\circ} \cdot \sin 27^{\circ}}$
$=\frac{2}{2 \cos 9^{\circ} \cdot \sin 9^{\circ}}-\frac{2}{2 \cos 27^{0} \cdot \sin 27^{\circ}}$
$=\frac{2}{\sin 18^{0}}-\frac{2}{\sin 54^{\circ}}=\frac{2 \cdot\left(\sin 54^{\circ}-\sin 18^{\circ}\right)}{\sin 54^{\circ} \cdot \sin 18^{\circ}}$
$=\frac{2.2 \cdot \cos 36^{\circ} \cdot \sin 18^{\circ}}{\sin 54^{\circ} \cdot \sin 18^{\circ}}=\frac{4 \cdot \cos 36^{\circ}}{\sin 54^{\circ}}$
$=\frac{4 \cdot \sin 54^{\circ}}{\sin 54^{\circ}}=4$
$\left(\right.$ vì $\left.\cos 36^{\circ}=\sin \left(90^{\circ}-36^{\circ}\right)=\sin 54^{\circ}\right)$