Question Number 21009 by xxyy@gmail.com last updated on 10/Sep/17 Answered by Tinkutara last updated on 10/Sep/17 $$\mathrm{2cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{5}\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{7}\pi}{\mathrm{16}} \\ $$$$=\left(\mathrm{4cos}\:\frac{\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{16}}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{4cos}\:\frac{\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{sin}\:\frac{\pi}{\mathrm{16}}\right)^{\mathrm{2}}…
Question Number 86534 by redmiiuser last updated on 29/Mar/20 $${Mr}.{Tanmay}\:{can}\:{you} \\ $$$${please}\:{help}\:{me}\:{in} \\ $$$${question}\:{no}.\mathrm{86454} \\ $$ Commented by mr W last updated on 29/Mar/20 $${it}'{s}\:{getting}\:{boring}……
Question Number 86532 by redmiiuser last updated on 29/Mar/20 $$\int\frac{{x}^{\mathrm{7}} .{dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\left(\mathrm{1}/\mathrm{2}\right)} } \\ $$ Commented by redmiiuser last updated on 29/Mar/20 $${Can}\:{anyone}\:{solve}\:{this} \\ $$$${in}\:{an}\:{easy}\:{way}.…
Question Number 20988 by aplus last updated on 09/Sep/17 Commented by aplus last updated on 09/Sep/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152001 by otchereabdullai@gmail.com last updated on 25/Aug/21 $$\mathrm{Four}\:\mathrm{members}\:\mathrm{of}\:\mathrm{a}\:\mathrm{school}'\mathrm{s}\:\mathrm{first}\:\mathrm{eleven} \\ $$$$\mathrm{criket}\:\mathrm{team}\:\mathrm{are}\:\mathrm{also}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{fourteen} \\ $$$$\mathrm{rugby}\:\mathrm{team}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{members}\:\mathrm{play} \\ $$$$\mathrm{for}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{teams} \\ $$ Answered by mr W last updated on…
Question Number 152000 by Olaf_Thorendsen last updated on 25/Aug/21 $$\bullet\:{Histoire}\:{de}\:{code}\:{secret}\: \\ $$$$ \\ $$$$\mathrm{Vous}\:\hat {\mathrm{e}tes}\:\mathrm{un}\:\mathrm{espion}\:\mathrm{et}\:\mathrm{vous}\:\mathrm{souhaitez} \\ $$$$\mathrm{assister}\:\:\grave {\mathrm{a}}\:\mathrm{une}\:\mathrm{r}\acute {\mathrm{e}union}\:\mathrm{secr}\grave {\mathrm{e}te}\:\mathrm{dans}\:\mathrm{une} \\ $$$$\mathrm{une}\:\mathrm{ambassade}\:\mathrm{ou}\:\mathrm{le}\:\mathrm{garde}\:\mathrm{demande} \\ $$$$\mathrm{un}\:\mathrm{mot}\:\mathrm{de}\:\mathrm{passe}… \\…
Question Number 151999 by otchereabdullai@gmail.com last updated on 24/Aug/21 $$\int\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{12}} \\ $$ Answered by Olaf_Thorendsen last updated on 25/Aug/21 $$\mathrm{F}\left({x}\right)\:=\:\int\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}−\mathrm{12}}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}−\mathrm{12}}−\frac{\mathrm{3}}{{x}^{\mathrm{2}}…
Question Number 86454 by redmiiuser last updated on 28/Mar/20 $$\:{pls}\:{check}\:{the}\: \\ $$$${question}\:{below} \\ $$ Answered by redmiiuser last updated on 28/Mar/20 $$\int\mathrm{1}/\sqrt{}\left(\mathrm{1}+{x}^{\mathrm{3}} \right).{dx} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{3}}…
Question Number 20876 by ajfour last updated on 05/Sep/17 Commented by ajfour last updated on 05/Sep/17 $${Rod}\:{AB}\:{is}\:{initially}\:{on}\:{a}\:{tabletop}. \\ $$$${End}\:{a}\:{given}\:{only}\:{a}\:{little}\:{push}\:{to} \\ $$$${start}\:{the}\:{fall}\:{of}\:{the}\:{rod}\:{as}\:{it}\:{turns} \\ $$$${through}\:{end}\:{B}\:{where}\:{static}\:{friction} \\ $$$${coefficient}\:{is}\:{high}\:{enough}.…
Question Number 151943 by Huy last updated on 24/Aug/21 $$\mathrm{Prove}\:\mathrm{arctan1}+\mathrm{arctan2}+\mathrm{arctan3}=\pi \\ $$ Commented by puissant last updated on 24/Aug/21 $${x}={arctan}\mathrm{1}\:,\:{y}={arctan}\mathrm{2}\:,\:{z}={arctan}\mathrm{3} \\ $$$${p}={x}+{y} \\ $$$$\Rightarrow\:{tan}\left({p}\right)={tan}\left({x}+{y}\right)=\frac{{tanx}+{tany}}{\mathrm{1}−{tanxtany}} \\…