Question Number 196567 by pete last updated on 27/Aug/23 $$\mathrm{Show}\:\mathrm{that}\:\left(\frac{\mathrm{1}+\:\mathrm{itan}\theta}{\mathrm{1}−\:\mathrm{itan}\theta}\right)^{\mathrm{n}} =\frac{\mathrm{1}+\:\mathrm{itan}\left(\mathrm{n}\theta\right)}{\mathrm{1}−\:\mathrm{tan}\left(\mathrm{n}\theta\right)} \\ $$ Commented by mokys last updated on 27/Aug/23 $$\left(\frac{\mathrm{1}+{itan}\theta}{\mathrm{1}−{itan}\theta}\right)^{{n}} =\left(\frac{\frac{{cos}\theta+{isin}\theta}{{cos}\theta}}{\frac{{cos}\theta−{isin}\theta}{{cos}\theta}}\right)^{{n}} =\left(\frac{{cos}\left({n}\theta\right)+{isin}\left({n}\theta\right)}{{cos}\left({n}\theta\right)−{isin}\left({n}\theta\right)}\right) \\ $$$$…
Question Number 196560 by peter frank last updated on 27/Aug/23 Commented by Spillover last updated on 27/Aug/23 Check your solution https://m.facebook.com/story.php?story_fbid=pfbid0QHNhoZjesQzGqy7jHAgbvRSRUx25yXHezsGoh2qHLCapd9KB4z5PFi88czsFYxZ6l&id=100084816875916&mibextid=Nif5oz Answered by MM42 last updated on 27/Aug/23…
Question Number 196525 by Mastermind last updated on 26/Aug/23 Answered by mr W last updated on 27/Aug/23 $$\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}>\sqrt{{m}^{\mathrm{2}} }={m} \\ $$$$\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}<\sqrt{{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}}=\sqrt{\left({m}+\mathrm{1}\right)^{\mathrm{2}} }={m}+\mathrm{1}…
Question Number 196444 by Mastermind last updated on 24/Aug/23 Commented by AST last updated on 25/Aug/23 $$=\left(\frac{\mathrm{6}+\mathrm{16}+\mathrm{6}+\mathrm{24}+\mathrm{20}+\mathrm{21}+\mathrm{144}}{\:\mathrm{24}}\right)×\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{79}}{\mathrm{56}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 196242 by pete last updated on 20/Aug/23 $$\mathrm{Simplify}\:\left(\frac{\mathrm{1}+\mathrm{cos2}\theta\:+\mathrm{isin2}\theta}{\mathrm{1}+\mathrm{cos2}\theta\:−\mathrm{isin2}\theta}\right)^{\mathrm{30}} \\ $$ Answered by MM42 last updated on 20/Aug/23 $$\left(\frac{\mathrm{2}{cos}^{\mathrm{2}} \theta+\mathrm{2}{isin}\theta{cos}\theta}{\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{2}{isin}\theta{cos}\theta}\right)^{\mathrm{30}} =\left(\frac{{cos}\theta+{isin}\theta}{{cos}\theta−{isin}\theta}\right)^{\mathrm{30}} \\ $$$$={cos}\left(\mathrm{60}\theta\right)+{isin}\left(\mathrm{60}\theta\right)\:\checkmark…
Question Number 196211 by pete last updated on 20/Aug/23 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}+\mathrm{i}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{e}+\frac{\mathrm{1}}{\mathrm{e}}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{e}−\frac{\mathrm{1}}{\mathrm{e}}\right)\mathrm{i} \\ $$ Answered by Frix last updated on 20/Aug/23 $$\mathrm{cos}\:{x}\:=\frac{\mathrm{e}^{\mathrm{2i}{x}} +\mathrm{1}}{\mathrm{2e}^{\mathrm{i}{x}} } \\ $$$$\mathrm{cos}\:\left({a}+{b}\mathrm{i}\right)\:=\mathrm{cos}\:{a}\:\mathrm{cosh}\:{b}\:−\mathrm{i}\:\mathrm{sin}\:{a}\:\mathrm{sinh}\:{b} \\…
Question Number 196205 by Mastermind last updated on 19/Aug/23 $$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{value}\:\mathrm{for}\:!\mathrm{7}\:? \\ $$ Answered by AST last updated on 19/Aug/23 $$!\mathrm{7}=\mathrm{7}!\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}+\frac{\mathrm{1}}{\mathrm{6}!}−\frac{\mathrm{1}}{\mathrm{7}!}\right)=\mathrm{1854} \\ $$ Commented by AST…
Question Number 195948 by Mastermind last updated on 13/Aug/23 $$\mathrm{An}\:\mathrm{object}\:\mathrm{is}\:\mathrm{said}\:\mathrm{to}\:\mathrm{cross}\:\mathrm{a}\:\mathrm{thousand}\: \\ $$$$\mathrm{kilimeters}\:\mathrm{in}\:\mathrm{planks}\:\mathrm{constant}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{times}\:\mathrm{faster}\:\mathrm{is}\:\mathrm{the}\:\mathrm{object}\:\mathrm{to}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of} \\ $$$$\mathrm{light}.\:\mathrm{plank}'\mathrm{s}\:\mathrm{time}\:=\:\mathrm{10}^{−\mathrm{44}} \mathrm{sec}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 195848 by pete last updated on 11/Aug/23 $$\mathrm{Please}\:\mathrm{how}\:\mathrm{did}\:\mid\mathrm{z}−\mathrm{a}\mid=\mathrm{r}\:\mathrm{became}\: \\ $$$$\mathrm{z}=\:\mathrm{a}\:+\:\mathrm{re}^{\mathrm{i}\theta} ? \\ $$ Answered by Frix last updated on 11/Aug/23 $$\mid{z}−{a}\mid={r}\:\Rightarrow\:{z}−{a}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{This}\:\mathrm{step}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true},\:\mathrm{even}\:\mathrm{for}\:{z}−{a}\in\mathbb{R}…
Question Number 195747 by Mastermind last updated on 09/Aug/23 Commented by AST last updated on 09/Aug/23 $${The}\:{total}\:{amount}\:{spent}\:{does}\:{not}\:{have}\:{to}\:{be} \\ $$$${necessarily}\:{equal}\:{to}\:{the}\:{sum}\:{of}\:{the}\:{remaining} \\ $$$${amount}\:{at}\:{each}\:{step}\:{of}\:{your}\:{spending}.\:{In}\:{fact}, \\ $$$${we}\:{can}\:{make}\:{that}\:{N}\mathrm{51}\:{grow}\:{arbitrarily}\:{large} \\ $$$${by}\:{spending}\:\epsilon\left({such}\:{as}\:\mathrm{0}.\mathrm{000000001}\right)\:{at}\:{every}\:{step}.…