Question Number 135608 by 0731619177 last updated on 14/Mar/21 Answered by dhgt last updated on 05/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 70075 by ajfour last updated on 30/Sep/19 Commented by behi83417@gmail.com last updated on 30/Sep/19 Commented by behi83417@gmail.com last updated on 30/Sep/19 $$\measuredangle\mathrm{ACB}=\frac{\mathrm{540}}{\mathrm{5}}=\mathrm{108}^{\bullet} \\…
Question Number 135610 by mnjuly1970 last updated on 14/Mar/21 $$\:\:\:\:\:\:\:\:\:\:….\:\mathscr{A}{dvanced}\:\:……\:\:\mathscr{C}{alculus}…. \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:{that}\:: \\ $$$$\:\:\:\begin{array}{|c|c|}{{i}\:::\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:={cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:\:\:\:\checkmark\:\:}\\{{ii}\:::\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)=\:{cosh}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:\checkmark\checkmark}\\\hline\end{array}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………. \\…
Question Number 4535 by FilupSmith last updated on 05/Feb/16 $$\mathrm{Lets}\:\mathrm{say}\:\mathrm{we}\:\mathrm{have}\:\mathrm{three}\:\mathrm{points}: \\ $$$${A}\left(\mathrm{0},\:\mathrm{0}\right) \\ $$$${B}\left({x},\:{y}\right) \\ $$$${C}\left(\delta{x},\:\delta{y}\right) \\ $$$$ \\ $$$$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{both}\:{B}\:\mathrm{and}\:{C}\:\mathrm{are}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{fuction}\:{y}={f}\left({x}\right),\:\mathrm{we}\:\mathrm{can}\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{under}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where}\:\mathrm{it}\:\mathrm{makes} \\…
Question Number 135604 by liberty last updated on 14/Mar/21 $$\:\frac{\mathrm{sin}\:\left(\pi/\mathrm{7}\right).\mathrm{cos}\:\left(\pi/\mathrm{14}\right)}{\mathrm{tan}\:\left(\mathrm{3}\pi/\mathrm{14}\right)\left(\mathrm{2cos}\:\left(\pi/\mathrm{7}\right)−\mathrm{1}\right)}\:=? \\ $$ Answered by EDWIN88 last updated on 14/Mar/21 $$\:\mathrm{Remark}\:\begin{cases}{\mathrm{sin}\:\mathrm{3x}=\mathrm{3sin}\:\mathrm{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{x}}\\{\mathrm{cos}\:\mathrm{3x}=\mathrm{4cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3cos}\:\mathrm{x}}\end{cases} \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\frac{\pi}{\mathrm{14}}\:\Rightarrow\frac{\mathrm{sin}\:\left(\pi/\mathrm{7}\right)\mathrm{cos}\:\left(\pi/\mathrm{14}\right)}{\mathrm{tan}\:\left(\mathrm{3}\pi/\mathrm{14}\right)\left(\mathrm{2cos}\:\left(\pi/\mathrm{7}\right)−\mathrm{1}\right)}\: \\…
Question Number 70069 by tw000001 last updated on 01/Oct/19 $$\underset{{n}=\mathrm{1}} {\overset{\mathrm{5}} {\prod}}\frac{\left(\mathrm{12}{n}−\mathrm{2}\right)^{\mathrm{4}} +\mathrm{18}^{\mathrm{2}} }{\left(\mathrm{12}{n}−\mathrm{8}\right)^{\mathrm{4}} +\mathrm{18}^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{10}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{22}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{34}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{46}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{58}^{\mathrm{4}} +\mathrm{324}\right)}{\left(\mathrm{4}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{16}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{28}^{\mathrm{4}}…
Question Number 70066 by Maclaurin Stickker last updated on 30/Sep/19 $${Solve} \\ $$$$\left.\mathrm{a}\right)\:{e}^{\mathrm{2}{x}} −{e}^{{x}+\mathrm{1}} −{e}^{{x}} +{e}<\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{4}.\mathrm{2}^{\mathrm{2}{x}} −\mathrm{9}.\mathrm{2}^{{x}} <−\mathrm{2} \\ $$$$\left.{c}\right)\mathrm{9}^{{x}} −\mathrm{4}.\mathrm{3}^{{x}+\mathrm{1}} +\mathrm{27}>\mathrm{0} \\…
Question Number 135603 by liberty last updated on 14/Mar/21 $$\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{17}}\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135602 by Algoritm last updated on 14/Mar/21 Commented by Dwaipayan Shikari last updated on 14/Mar/21 $${Transendental}\:{Numbers}\:{can}'{t}\:{be}\:{An}\:{Algebraic}\:{solution} \\ $$ Terms of Service Privacy Policy…
Question Number 135599 by liberty last updated on 14/Mar/21 Answered by mathDivergent last updated on 14/Mar/21 $$\mathrm{2}^{\mathrm{8}} ,\:\mathrm{I}\:\mathrm{have}\:\mathrm{blunder} \\ $$ Commented by EDWIN88 last updated…