Tính
a) \(\cos \dfrac{\pi }{9} + \cos \dfrac{{2\pi }}{9} + \ldots + \cos \dfrac{{8\pi }}{9};\)
b) \({\sin ^2}\dfrac{\pi }{3} + {\sin ^2}\dfrac{\pi }{6} + {\sin ^2}\dfrac{\pi }{9} + {\sin ^2}\dfrac{{2\pi }}{9} + {\sin ^2}\dfrac{{5\pi }}{{18}} + {\sin ^2}\dfrac{{7\pi }}{{18}}\);
c) \({\cos ^2}\dfrac{\pi }{3} + {\cos ^2}\dfrac{{5\pi }}{6} + {\cos ^2}\dfrac{\pi }{9} + {\cos ^2}\dfrac{{11\pi }}{{18}} + {\cos ^2}\dfrac{{13\pi }}{{18}} + {\cos ^2}\dfrac{{2\pi }}{9}\);
d) \(\cos \dfrac{\pi }{5} + \cos \dfrac{{2\pi }}{5} + \ldots + \cos \dfrac{{9\pi }}{5};\)
e) \(\sin \dfrac{\pi }{5} + \sin \dfrac{{2\pi }}{5} + \ldots + \sin \dfrac{{9\pi }}{5}\)
a)
\(\cos \dfrac{\pi }{9} + \cos \dfrac{{2\pi }}{9} + \ldots + \cos \dfrac{{8\pi }}{9} = 0\), do \(\cos \left( {\pi - \alpha } \right) = - \cos \alpha .\)
b) Do \(\sin \dfrac{\pi }{3} = \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{6}\) nên \({\sin ^2}\dfrac{\pi }{3} + {\sin ^2}\dfrac{\pi }{6} = 1.\)
Do \(\sin \dfrac{{7\pi }}{{18}} = \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{9}} \right) = \cos \dfrac{\pi }{9}\) nên \({\sin ^2}\dfrac{{7\pi }}{{18}} + {\sin ^2}\dfrac{\pi }{9} = 1\).
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Do \(\sin \dfrac{{5\pi }}{{18}} = \sin \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{9}} \right) = \cos \dfrac{{2\pi }}{9}\) nên \({\sin ^2}\dfrac{{2\pi }}{9} + {\sin ^2}\dfrac{{5\pi }}{{18}} = 1\).
Vậy \({\sin ^2}\dfrac{\pi }{3} + {\sin ^2}\dfrac{\pi }{6} + {\sin ^2}\dfrac{\pi }{9} + {\sin ^2}\dfrac{{2\pi }}{9} + {\sin ^2}\dfrac{{5\pi }}{{18}} + {\sin ^2}\dfrac{{7\pi }}{{18}} = 3\)
c) Do \(\cos \left( {\dfrac{{5\pi }}{6}} \right) = \cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{3}} \right) = - \sin \dfrac{\pi }{3}\), nên \({\cos ^2}\dfrac{\pi }{3} + {\cos ^2}\dfrac{{5\pi }}{6} = 1\).
Do \(\cos \dfrac{{11\pi }}{{18}} = \cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{9}} \right) = - \sin \dfrac{\pi }{9}\), nên \({\cos ^2}\dfrac{\pi }{9} + {\cos ^2}\dfrac{{11\pi }}{{18}} = 1\)
Do \(\cos \dfrac{{13\pi }}{{18}} = \cos \left( {\dfrac{\pi }{2} + \dfrac{{2\pi }}{9}} \right) = - \sin \dfrac{{2\pi }}{9}\), nên \({\cos ^2}\dfrac{{13\pi }}{{18}} + {\cos ^2}\dfrac{{2\pi }}{9} = 1\)
Vậy \({\cos ^2}\dfrac{\pi }{3} + {\cos ^2}\dfrac{{5\pi }}{6} + {\cos ^2}\dfrac{\pi }{9} + {\cos ^2}\dfrac{{11\pi }}{{18}} + {\cos ^2}\dfrac{{13\pi }}{{18}} + {\cos ^2}\dfrac{{2\pi }}{9} = 3\)
d) Do \(\cos \dfrac{{6\pi }}{5} = \cos \left( {\pi + \dfrac{\pi }{5}} \right) = - \cos \dfrac{\pi }{5};\) \(\cos \dfrac{{7\pi }}{5} = - \cos \dfrac{{2\pi }}{5};\cos \dfrac{{8\pi }}{5} = - \cos \dfrac{{3\pi }}{5};\) \(\cos \dfrac{{9\pi }}{5} = - \cos \dfrac{{4\pi }}{5};\cos \pi = - 1\) nên \(\cos \dfrac{\pi }{5} + \cos \dfrac{{2\pi }}{5} + \ldots + \cos \dfrac{{9\pi }}{5} = - 1\)
e) Tương tự đối với sin, nhưng ở đây \(\sin \pi = 0\), ta có :
\(\sin \dfrac{\pi }{5} + \sin \dfrac{{2\pi }}{5} + \ldots + \sin \dfrac{{9\pi }}{5} = 0.\)
(Chú ý: Ta cũng có thể xét thập giác đều có các đỉnh là \({A_k}\) là các điểm trên đường tròn lượng giác, xác định bởi các số \(\dfrac{{k\pi }}{5}\) (k = 1; 2; 3; 4; ....; 9; 10) và nhận xét rằng \(\overrightarrow {O{A_1}} + \overrightarrow {O{A_2}} + \ldots \overrightarrow {O{A_{10}}} = \overrightarrow 0 \))