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Question-219839

Question Number 219839 by Lekhraj last updated on 02/May/25 Answered by golsendro last updated on 03/May/25 $$\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{2}\left(\frac{\sqrt{\mathrm{4}−\sqrt{\mathrm{5}}}\:+\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\mathrm{2}}\:\right)\left(\frac{\sqrt{\mathrm{4}−\sqrt{\mathrm{5}}}−\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\mathrm{2}}\:\right) \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}−\sqrt{\mathrm{5}}\:−\sqrt{\mathrm{5}}\:+\mathrm{2}\:\right) \\ $$$$\:=\:\mathrm{3}−\sqrt{\mathrm{5}}\: \\ $$ Terms of…

lim-n-n-1-1-n-1-2-n-1-2n-ln-2-

Question Number 219832 by mnjuly1970 last updated on 02/May/25 $$ \\ $$$$\:\:\:\:\:{lim}_{{n}\rightarrow\infty} \:{n}\left(\frac{\mathrm{1}}{\mathrm{1}+{n}}\:+\frac{\mathrm{1}}{\mathrm{2}+{n}}\:+…+\frac{\mathrm{1}}{\mathrm{2}{n}}\:−{ln}\left(\mathrm{2}\right)\right)=? \\ $$$$ \\ $$ Answered by universe last updated on 03/May/25 Commented…