Question Number 93256 by i jagooll last updated on 12/May/20 $$\begin{cases}{\mathrm{3}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{4}\:\mathrm{cos}\:\mathrm{y}}\\{\mathrm{3}\:\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{4sin}\:\mathrm{y}\:=\:\mathrm{5}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{x}\:\&\mathrm{y}\:\mathrm{with}\:\mathrm{acute}\:\mathrm{angle}\: \\ $$ Commented by john santu last updated on 12/May/20 $$\begin{cases}{\mathrm{9}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{16}\:\mathrm{cos}\:^{\mathrm{2}}…
Question Number 27706 by shivamgiri last updated on 13/Jan/18 $${the}\:{number}\:{of}\:{x}\epsilon\left[\mathrm{0},\mathrm{2}\pi\right]{for}\:{which}\mid\sqrt{}\mathrm{2}{sin}^{\mathrm{4}\:} {x}+\mathrm{18}{cos}^{\mathrm{2}} {x}−\sqrt{\mathrm{2}{cos}^{\mathrm{4}} {x}+\mathrm{18}{sin}^{\mathrm{2}} {x}}\mid=\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 93239 by Ar Brandon last updated on 11/May/20 $$\mathrm{Calculate}; \\ $$$$\left.{i}\right)\:\mathrm{cos}\left(\mathrm{arctan}\:{x}\right) \\ $$$$\left.{ii}\right)\:\mathrm{cos}\left(\mathrm{arcsin}\:{x}\right) \\ $$$$\left.{iii}\right)\:\mathrm{tan}\left(\mathrm{arcsin}\:{x}\right) \\ $$ Commented by mathmax by abdo last…
Question Number 27640 by Tinkutara last updated on 11/Jan/18 Answered by ajfour last updated on 11/Jan/18 $$\mathrm{cos}\:{C}=\frac{\mathrm{3}}{\mathrm{5}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\Rightarrow\:\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\left({n}+\mathrm{2}\right)^{\mathrm{2}} −{n}^{\mathrm{2}} }{\mathrm{2}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{\mathrm{3}}{\mathrm{5}}…
Question Number 27605 by ajfour last updated on 10/Jan/18 Answered by mrW2 last updated on 13/Jan/18 $$\frac{{DB}}{{DC}}=\frac{{AB}}{{AC}}=\frac{{c}}{{b}} \\ $$$$\Rightarrow{DB}=\frac{{ac}}{{b}+{c}} \\ $$$$\Rightarrow{DC}=\frac{{ab}}{{b}+{c}} \\ $$$${AD}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{b}+{c}} \\ $$$$\frac{{AE}}{{AB}}=\frac{{AD}}{{DB}}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{b}+{c}}×\frac{{b}+{c}}{{ac}}=\frac{\mathrm{2}\sqrt{{s}\left({s}−{a}\right){bc}}}{{ac}}…
Question Number 158523 by gsk2684 last updated on 05/Nov/21 $${find}\:{the}\:{number}\:{of}\:{values}\:{of}\:{p} \\ $$$${for}\:{which}\:{equation}\: \\ $$$$\mathrm{sin}^{\mathrm{3}} {x}+\mathrm{1}+{p}^{\mathrm{3}} −\mathrm{3}{p}\:\mathrm{sin}\:{x}\:=\mathrm{0}\left({p}>\mathrm{0}\right) \\ $$$${has}\:{a}\:{root}? \\ $$ Answered by mr W last…
Question Number 27393 by kumar123 last updated on 06/Jan/18 $${solve} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}/\mathrm{1}−{x}^{\mathrm{2}} \right)+\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} /\mathrm{2}{x}\right)=\pi/\mathrm{3} \\ $$ Answered by $@ty@m last updated on 06/Jan/18…
Question Number 158358 by mnjuly1970 last updated on 03/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158354 by cortano last updated on 03/Nov/21 $$\:\sqrt{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+{x}\right)}\:+\sqrt{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)}\:=\sqrt[{\mathrm{4}}]{\mathrm{2cos}\:\mathrm{2}{x}} \\ $$ Commented by tounghoungko last updated on 03/Nov/21 $$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+{x}\right)+\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)+\mathrm{2}\sqrt{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+{x}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)}=\sqrt{\mathrm{2cos}\:\mathrm{2}{x}} \\ $$$$\left({i}\right)\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+{x}\right)+\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)=\mathrm{2sin}\:\frac{\pi}{\mathrm{4}}\mathrm{cos}\:{x}=\sqrt{\mathrm{2}}\:\mathrm{cos}\:{x} \\ $$$$\left({ii}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+{x}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}−\mathrm{cos}\:\mathrm{2}{x}\right)=\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}} \\…
Question Number 92781 by i jagooll last updated on 09/May/20 $$\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)\: \\ $$ Commented by john santu last updated on 09/May/20 $$\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}}…