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Bài 9. Tính
a) \(4(cos{24^0} + \cos {48^0} – \cos {84^0} – \cos {12^0})\)
b) \(96\sqrt 3 \sin {\pi \over {48}}\cos {\pi \over {48}}\cos {\pi \over {24}}\cos {\pi \over {12}}\cos {\pi \over 6}\)
c) \(\tan {9^0} – \tan {63^0} + \tan {81^0} – \tan {27^0}\)
a)
\(\eqalign{
& cos{24^0} + \cos {48^0} = \cos ({36^0} – {12^0}) + \cos ({36^0} + {12^0}) \cr
& = 2\cos {36^0}\cos {12^0} \cr
& \cos {84^0} + \cos {12^0} = 2\cos {36^0}\cos {48^0} \cr
& 4(\cos {24^0} + \cos {48^0} – \cos {84^0} – \cos {12^0}) = 8\cos {36^0}(\cos {12^0} – \cos {48^0}) \cr
& = 8\cos {36^0}.2\sin {30^0}.\sin {18^0} = 8\cos {36^0}\sin {18^0} \cr
& = 8\cos {36^0}.\sqrt {{{1 – \cos {{36}^0}} \over 2}} \cr} \)
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 :
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\(4(cos{24^0} + \cos {48^0} – \cos {84^0} – \cos {12^0}) = 2(1 + \sqrt 5 )\sqrt {{{3 – \sqrt 5 } \over 8}} = 2\)
b)
\(\eqalign{
& 96\sqrt 3 \sin {\pi \over {48}}\cos {\pi \over {48}}\cos {\pi \over {24}}\cos {\pi \over {12}}\cos {\pi \over 6} \cr
& = 48\sqrt 3 \sin {\pi \over {24}}\cos {\pi \over {24}}\cos {\pi \over {12}}\cos {\pi \over 6} \cr
& = 24\sqrt 3 \sin {\pi \over {12}}\cos {\pi \over {12}}\cos {\pi \over 6} \cr
& = 12\sqrt 3 \sin {\pi \over 6}\cos {\pi \over 6} = 6\sqrt 3 \sin {\pi \over 3} = 9 \cr} \)
c)
\(\eqalign{
& \tan {9^0} – \tan {63^0} + \tan {81^0} – \tan {27^0} \cr
& = {{\cos {{81}^0}} \over {\sin {{81}^0}}} + {{\sin {{81}^0}} \over {\cos {{81}^0}}} – ({{\cos {{27}^0}} \over {\sin {{27}^0}}} + {{\sin {{27}^0}} \over {\cos {{27}^0}}}) \cr
& = {1 \over {\sin {{81}^0}.cos{{81}^0}}} – {1 \over {\sin {{27}^0}.cos{{27}^0}}} \cr
& = {2 \over {\sin {{18}^0}}} – {2 \over {\sin {{54}^0}}} = {2 \over {\cos {{72}^0}}} – {2 \over {\cos {{36}^0}}} \cr
& = {2 \over {2{{\cos }^2}{{36}^0} – 1}} – {2 \over {\cos {{36}^0}}} \cr} \)
Thay \(\cos {36^0} = {{1 + \sqrt 5 } \over 4}\) ta được: \(\tan {9^0} – \tan {63^0} + \tan {81^0} – \tan {27^0} = 4\)