Câu hỏi của Ngô Thành Chung

Question

Tính A = $\dfrac{1}{cos\dfrac{\pi}{7}}+\dfrac{1}{cos\dfrac{3\pi}{7}}+\dfrac{1}{cos\dfrac{5\pi}{7}}$

Lê Thị Thục Hiền · Accepted Answer

A$=\dfrac{cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}}{cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{7}.cos\dfrac{5\pi}{7}}$Đặt tử là Y; mẫu là UCó $Y=$$cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+\left(cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}\right)$$=cos\left(\pi-\dfrac{2\pi}{7}\right).cos\left(\pi-\dfrac{4\pi}{7}\right)+cos\dfrac{\pi}{7}\left(cos\dfrac{5\pi}{7}+cos\dfrac{3\pi}{7}\right)$$=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{\pi}{7}.2cos\dfrac{4\pi}{7}.cos\dfrac{\pi}{7}$$=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+2.cos^2\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}$$=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+\left(cos\dfrac{2\pi}{7}+1\right).cos\dfrac{4\pi}{7}$$=2.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{4\pi}{7}$$=cos\dfrac{6\pi}{7}+cos\dfrac{2\pi}{7}+cos\dfrac{4\pi}{7}$$\Rightarrow sin\dfrac{\pi}{7}.Y=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}$$=\dfrac{1}{2}\left(-sin\dfrac{\pi}{7}+sin\dfrac{3\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{3\pi}{7}+sin\dfrac{5\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{5\pi}{7}+sin\pi\right)$$=\dfrac{1}{2}\left(sin\pi-sin\dfrac{\pi}{7}\right)$$=-\dfrac{1}{2}sin\dfrac{\pi}{7}$$\Rightarrow Y=-\dfrac{1}{2}$Có $sin\dfrac{\pi}{7}.U=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{5}.cos\dfrac{5\pi}{7}$$=\dfrac{1}{2}.sin\dfrac{2\pi}{7}.cos\left(\pi-\dfrac{2\pi}{7}\right).cos\dfrac{3\pi}{5}$$=-\dfrac{1}{4}.sin\dfrac{4\pi}{7}.cos\left(\pi-\dfrac{4\pi}{5}\right)$$=\dfrac{1}{8}.sin\dfrac{8\pi}{7}$$=\dfrac{1}{8}.sin\left(\pi+\dfrac{\pi}{7}\right)=-\dfrac{1}{8}.sin\dfrac{\pi}{7}$$\Rightarrow U=-\dfrac{1}{8}$ Vậy $A=\dfrac{Y}{U}=4$Câu trả lời: A$=\dfrac{cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}}{cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{7}.cos\dfrac{5\pi}{7}}$Đặt tử là Y; mẫu là UCó $Y=$$cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+\left(cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}\right)$$=cos\left(\pi-\dfrac{2\pi}{7}\right).cos\left(\pi-\dfrac{4\pi}{7}\right)+cos\dfrac{\pi}{7}\left(cos\dfrac{5\pi}{7}+cos\dfrac{3\pi}{7}\right)$$=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{\pi}{7}.2cos\dfrac{4\pi}{7}.cos\dfrac{\pi}{7}$$=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+2.cos^2\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}$$=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+\left(cos\dfrac{2\pi}{7}+1\right).cos\dfrac{4\pi}{7}$$=2.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{4\pi}{7}$$=cos\dfrac{6\pi}{7}+cos\dfrac{2\pi}{7}+cos\dfrac{4\pi}{7}$$\Rightarrow sin\dfrac{\pi}{7}.Y=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}$$=\dfrac{1}{2}\left(-sin\dfrac{\pi}{7}+sin\dfrac{3\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{3\pi}{7}+sin\dfrac{5\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{5\pi}{7}+sin\pi\right)$$=\dfrac{1}{2}\left(sin\pi-sin\dfrac{\pi}{7}\right)$$=-\dfrac{1}{2}sin\dfrac{\pi}{7}$$\Rightarrow Y=-\dfrac{1}{2}$Có $sin\dfrac{\pi}{7}.U=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{5}.cos\dfrac{5\pi}{7}$$=\dfrac{1}{2}.sin\dfrac{2\pi}{7}.cos\left(\pi-\dfrac{2\pi}{7}\right).cos\dfrac{3\pi}{5}$$=-\dfrac{1}{4}.sin\dfrac{4\pi}{7}.cos\left(\pi-\dfrac{4\pi}{5}\right)$$=\dfrac{1}{8}.sin\dfrac{8\pi}{7}$$=\dfrac{1}{8}.sin\left(\pi+\dfrac{\pi}{7}\right)=-\dfrac{1}{8}.sin\dfrac{\pi}{7}$$\Rightarrow U=-\dfrac{1}{8}$ Vậy $A=\dfrac{Y}{U}=4$

Etermintrude💫 · Answer

CHÚC BẠN HỌC TỐT NHÉCâu trả lời: CHÚC BẠN HỌC TỐT NHÉ

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