Menu Close

Question-130092




Question Number 130092 by Adel last updated on 22/Jan/21
Commented by 0731619177 last updated on 24/Jan/21
hhhh
$${hhhh} \\ $$
Answered by Dwaipayan Shikari last updated on 22/Jan/21
log(−((2/( (√2))).(2/( (√(2+(√2))))).(2/( (√(2+(√(2+(√2)))))))...)^3 )^1   cos(x/2)cos(x/4)cos(x/8)...=((sinx)/x)   (x=(π/2))  ⇒((√2)/2).((√(2+(√2)))/2).((√(2+(√(2+(√2)))))/2)...=(2/π)  =log((−(π/2))^3 )=iπ+3log((π/2))
$${log}\left(−\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}.\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}.\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}}…\right)^{\mathrm{3}} \right)^{\mathrm{1}} \\ $$$${cos}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{4}}{cos}\frac{{x}}{\mathrm{8}}…=\frac{{sinx}}{{x}}\:\:\:\left({x}=\frac{\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}}{\mathrm{2}}…=\frac{\mathrm{2}}{\pi} \\ $$$$={log}\left(\left(−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} \right)={i}\pi+\mathrm{3}{log}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *