Chứng minh: $cos\dfrac{\pi}{11} + cos\dfrac{3\pi}{11} + … + cos\dfrac{9\pi}{11}$

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1. Ta cminh

   $\cos \left( \dfrac{\pi}{11} \right) + \cos \left( \dfrac{3\pi}{11} \right) + \cdots + \cos \left( \dfrac{9\pi}{11} \right) = \dfrac{1}{2}$

   $<-> 2\cos \left( \dfrac{\pi}{11} \right) + 2\cos \left( \dfrac{3\pi}{11} \right) + \cdots + 2\cos \left( \dfrac{9\pi}{11} \right) = 1$

   Nhân cả hai vế với $\sin \left( \dfrac{\pi}{11} \right)$ ta cần cminh

   $2\cos \left( \dfrac{\pi}{11} \right) \sin \left( \dfrac{\pi}{11} \right) + 2\cos \left( \dfrac{3\pi}{11} \right) \sin \left( \dfrac{\pi}{11} \right) + \cdots + 2\cos \left( \dfrac{9\pi}{11} \right) \sin \left( \dfrac{\pi}{11} \right) = \sin \left( \dfrac{\pi}{11} \right)$

   Ta có

   $VT = 2\cos \left( \dfrac{\pi}{11} \right) \sin \left( \dfrac{\pi}{11} \right) + 2\cos \left( \dfrac{3\pi}{11} \right) \sin \left( \dfrac{\pi}{11} \right) + \cdots + 2\cos \left( \dfrac{9\pi}{11} \right) \sin \left( \dfrac{\pi}{11} \right)$

   Áp dụng biến đổi tích thành tổng ta có

   $VT = \sin \left( \dfrac{2\pi}{11} \right) + \sin \left( \dfrac{4\pi}{11} \right) – \sin\left( \dfrac{2\pi}{11} \right) + \sin \left( \dfrac{6\pi}{11}  \right) – \sin \left( \dfrac{4\pi}{11} \right) + \cdots + \sin \left( \dfrac{10\pi}{11} \right) – \sin\left( \dfrac{8\pi}{11} \right)$

   $= \sin \left( \dfrac{10\pi}{11} \right)$

   $= \sin \left(  \pi – \dfrac{\pi}{11} \right)$

   $= \sin \left( \dfrac{\pi}{11} \right) = VP$

   Vậy ta có đpcm

   Trả lời