Question Number 21116 by mondodotto@gmail.com last updated on 13/Sep/17 Answered by Tinkutara last updated on 13/Sep/17 $$\mathrm{tan}\:\mathrm{A}=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow\mathrm{sin}\:\mathrm{A}=\frac{\mathrm{3}}{\mathrm{5}},\mathrm{cos}\:\mathrm{A}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\mathrm{A}=\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{10}}\Rightarrow\mathrm{cos}\:\frac{\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$$\therefore\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}…
Question Number 86634 by liki last updated on 29/Mar/20 Answered by MJS last updated on 29/Mar/20 $$\int\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{tan}\:\theta\:−\mathrm{1}}{d}\theta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\theta\:\rightarrow\:{d}\theta=\mathrm{cos}^{\mathrm{2}} \:\theta\:{dt}\right] \\ $$$$=\int\frac{{dt}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=…
Question Number 21038 by ajfour last updated on 10/Sep/17 Commented by ajfour last updated on 10/Sep/17 $${Find}\:{angular}\:{velocity}\:{and}\:{angular} \\ $$$${acceleration}\:{of}\:{rod}\:{if}\:{end}\:{A}\:{moves} \\ $$$${to}\:{the}\:{right}\:{with}\:{constant}\:{velocity} \\ $$$${v}. \\ $$…
Question Number 152101 by otchereabdullai@gmail.com last updated on 25/Aug/21 $$\mathrm{Mr}\:\mathrm{Bonsu}\:\mathrm{an}\:\mathrm{engineer}\:\mathrm{walked}\:\mathrm{round} \\ $$$$\mathrm{a}\:\mathrm{cylindrical}\:\mathrm{container}\:\mathrm{4m}\:\mathrm{high}\:\mathrm{once} \\ $$$$\mathrm{keeping}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{1m}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{container}.\:\mathrm{If}\:\mathrm{he}\:\mathrm{walked}\:\mathrm{with}\:\mathrm{a}\: \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{3kmh}^{−\mathrm{1}} \:\mathrm{for}\:\mathrm{three}\:\mathrm{minutes},\: \\ $$$$\mathrm{calculate}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{whole}\:\mathrm{number} \\ $$$$\mathrm{the}: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}…
Question Number 86560 by naka3546 last updated on 29/Mar/20 $${x}\:\:=\:\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{15}}{\mathrm{32}}\:+\:\frac{\mathrm{105}}{\mathrm{384}}\:+\:… \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{1}\:\:=\:\:? \\ $$ Commented by redmiiuser last updated on 29/Mar/20 $${ans}\:{is}\:\mathrm{7}. \\ $$…
Question Number 86552 by M±th+et£s last updated on 29/Mar/20 $${proe}\:{that} \\ $$$${f}\left({x}\right)={x}+\left[{x}\right]\: \\ $$$${increase}\:{in}\:{R} \\ $$ Commented by mathmax by abdo last updated on 29/Mar/20…
Question Number 21015 by xxyy@gmail.com last updated on 10/Sep/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21013 by aplus last updated on 10/Sep/17 Commented by aplus last updated on 10/Sep/17 $$\mathrm{hhelp}\:\mathrm{guyselp}\:\mathrm{guys} \\ $$ Answered by Tinkutara last updated on…
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}……