Tính $A=cos\frac{\pi}{2^2}.cos\frac{\pi}{2^3}.cos\frac{\pi}{2^4}.....cos\frac{\pi}{2^{n+1}}$
Edited by Element hero Neos, 20-07-2017 - 20:21.
#2
Posted 26-07-2017 - 08:45
Math Master
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Blue Sky
Tính $A=cos\frac{\pi}{2^2}.cos\frac{\pi}{2^3}.cos\frac{\pi}{2^4}.....cos\frac{\pi}{2^{n+1}}$
Ta có $2^n.Sin \frac{\pi}{2^{n+1}} .A = 2^{n-1}.cos\frac{\pi}{2^2}...cos\frac{\pi}{2^n}.Sin\frac{\pi}{2^n} = cos\frac{\pi}{2} = 0 \Rightarrow A = ...$
Edited by Math Master, 26-07-2017 - 08:46.
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#3
Posted 26-07-2017 - 19:22
Element hero Neos
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Trung úy
Ta có $2^n.Sin \frac{\pi}{2^{n+1}} .A = 2^{n-1}.cos\frac{\pi}{2^2}...cos\frac{\pi}{2^n}.Sin\frac{\pi}{2^n} = cos\frac{\pi}{2} = 0 \Rightarrow A = ...$
Chỗ cuối cùng là $sin\frac{\pi}{2}$ chứ nhỉ?
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