Chứng minh:
a) \(\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7} = \dfrac{1}{2}\cot \dfrac{\pi }{{14}};\)
b) \(\cos \dfrac{\pi }{{11}} + \cos \dfrac{{3\pi }}{{11}} + \cos \dfrac{{5\pi }}{{11}} + \cos \dfrac{{7\pi }}{{11}} + \cos \dfrac{{9\pi }}{{11}}\)
\(= \dfrac{1}{2}\)
c) \(\cos \dfrac{{2\pi }}{{11}} + \cos \dfrac{{4\pi }}{{11}} + \cos \dfrac{{6\pi }}{{11}} + \cos \dfrac{{8\pi }}{{11}} + \cos \dfrac{{10\pi }}{{11}}\)
\(= - \dfrac{1}{2}\)
d) \(\sin \dfrac{\pi }{{11}} + \sin \dfrac{{2\pi }}{{11}} + \ldots + \sin \dfrac{{10\pi }}{{11}} = \cot \dfrac{\pi }{{22}}.\)
a) Ta có
\(\begin{array}{l}\sin \dfrac{{2\pi }}{7}\sin \dfrac{\pi }{7} = \dfrac{1}{2}\left( {\cos \dfrac{\pi }{7} - \cos \dfrac{{3\pi }}{7}} \right),\\\sin \dfrac{{4\pi }}{7}\sin \dfrac{\pi }{7} = \dfrac{1}{2}\left( {\cos \dfrac{{3\pi }}{7} - \cos \dfrac{{5\pi }}{7}} \right),\\\sin \dfrac{{6\pi }}{7}\sin \dfrac{\pi }{7} = \dfrac{1}{2}\left( {\cos \dfrac{{5\pi }}{7} - \cos \pi } \right)\end{array}\)
Từ đó
\(\begin{array}{l}\left( {\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}} \right)\sin \dfrac{\pi }{7}\\ = \dfrac{1}{2}\left( {1 + \cos \dfrac{\pi }{7}} \right) = {\cos ^2}\dfrac{\pi }{{14}}\end{array}\)
Do \(\sin \dfrac{\pi }{7} = 2\sin \dfrac{\pi }{{14}}\cos \dfrac{\pi }{{14}},\) ta suy ra
\(\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7} = \dfrac{1}{2}\cot \dfrac{\pi }{{14}}.\)
b) Với \(k = 1,2,3,4,5\) ta có:
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\(\cos \dfrac{{\left( {2k - 1} \right)\pi }}{{11}}\sin \dfrac{\pi }{{11}}\)
\(= \dfrac{1}{2}\left[ {\sin \dfrac{{2k\pi }}{{11}} - \sin \dfrac{{\left( {2k - 2} \right)\pi }}{{11}}} \right]\),
nên nếu gọi B là vế trái của đẳng thức ở câu b) thì
\(\begin{array}{l}B\sin \dfrac{\pi }{{11}} \\= \dfrac{1}{2}\left[ {\left( {\sin \dfrac{{2\pi }}{{11}} - \sin 0} \right) + \left( {\sin \dfrac{{4\pi }}{{11}} - \sin \dfrac{{2\pi }}{{11}}} \right)} \right.\\\left. { + \ldots + \left( {\sin \dfrac{{10\pi }}{{11}} - \sin \dfrac{{8\pi }}{{11}}} \right)} \right]\\ = \dfrac{1}{2}\sin \dfrac{{10\pi }}{{11}} = \dfrac{1}{2}\sin \dfrac{\pi }{{11}}.\end{array}\)
Từ đó \(B = \dfrac{1}{2}.\)
c) Với \(k = 1,2,3,4,5\) ta có
\(\cos \dfrac{{2k\pi }}{{11}}\sin \dfrac{\pi }{{11}}\)
\(= \dfrac{1}{2}\left[ {\sin \dfrac{{\left( {2k + 1} \right)\pi }}{{11}} - \sin \dfrac{{\left( {2k - 1} \right)\pi }}{{11}}} \right]\) nên gọi C là vế trái của đẳng thức câu c) thì
\(\begin{array}{l}C\sin \dfrac{\pi }{{11}} \\= \dfrac{1}{2}\left[ {\left( {\sin \dfrac{{3\pi }}{{11}} - \sin \dfrac{\pi }{{11}}} \right) + \left( {\sin \dfrac{{5\pi }}{{11}} - \sin \dfrac{{3\pi }}{{11}}} \right)} \right.\\ + \ldots + \left. {\left( {\sin \pi - \sin \dfrac{{9\pi }}{{11}}} \right)} \right]\\ = - \dfrac{1}{2}\sin \dfrac{\pi }{{11}}.\end{array}\)
Từ đó \(C = - \dfrac{1}{2}.\)
d) Gọi D là vế trái của bất đẳng thức câu d thì (ở đây \(n = 10,\alpha = \dfrac{\pi }{{11}}\))
\(D\sin \dfrac{\pi }{{22}} = \sin \dfrac{{10\pi }}{{22}}\sin \dfrac{\pi }{2} = \sin \dfrac{{10\pi }}{{22}} = \cos \dfrac{\pi }{{22}}\)
Từ đó \(D = \cot \dfrac{\pi }{{22}}.\)