Question Number 82944 by jagoll last updated on 26/Feb/20 $$ \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{27}^{\mathrm{o}} \right)+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{87}^{\mathrm{o}} \right)+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{33}^{\mathrm{o}} \right)\:= \\ $$ Commented by john santu last…
Question Number 17348 by Tinkutara last updated on 04/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{points}\:\mathrm{in}\:\left(−\infty,\:\infty\right)\:\mathrm{for} \\ $$$$\mathrm{which}\:{x}^{\mathrm{2}} \:−\:{x}\:\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:\mathrm{0}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{6} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{4} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{0} \\ $$ Answered by…
Question Number 82879 by jagoll last updated on 25/Feb/20 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{a}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{b}}{\mathrm{sin}\:\mathrm{acos}\:\mathrm{a}\:+\:\mathrm{sin}\:\mathrm{bcos}\:\mathrm{b}}\:=\:\mathrm{cot}\:\left(\mathrm{a}+\mathrm{b}\right) \\ $$ Commented by mahdi last updated on 25/Feb/20 $$\mathrm{problem}: \\…
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Question Number 17233 by sushmitak last updated on 02/Jul/17 $$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{out}\:\mathrm{if} \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)>\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{cos}\:\left(\mathrm{cos}\:{x}\right)\right)>\mathrm{sin}\:\left(\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\right) \\ $$$$\mathrm{exam}\:\mathrm{questions}. \\ $$$$\mathrm{calculators}\:\mathrm{not}\:\mathrm{allowed}. \\ $$ Commented…
Question Number 82767 by jagoll last updated on 24/Feb/20 $$\mathrm{If}\:\frac{\mathrm{sin}\:\left(\mathrm{A}+\theta\right)}{\mathrm{sin}\:\left({B}+\theta\right)}\:=\:\sqrt{\frac{\mathrm{sin}\:\mathrm{2}{A}}{\mathrm{sin}\:\mathrm{2}{B}}} \\ $$$${prove}\:{that}\:\mathrm{tan}\:^{\mathrm{2}} \theta\:=\:\mathrm{tan}\:{A}.\mathrm{tan}\:{B} \\ $$ Commented by jagoll last updated on 24/Feb/20 $${thank}\:{you}\:{mr}\:{w}\:{and}\:{john} \\ $$…
Question Number 82759 by jagoll last updated on 24/Feb/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2cos}\:\mathrm{80}^{{o}} }}}}\:=\: \\ $$ Answered by jagoll last updated on 24/Feb/20 Terms of Service…
Question Number 17153 by Tinkutara last updated on 01/Jul/17 $$\mathrm{If}\:{m},\:{n}\:\in\:{N}\left({n}\:>\:{m}\right),\:\mathrm{then}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${n}\mid\mathrm{sin}\:{x}\mid\:=\:{m}\mid\mathrm{sin}\:{x}\mid\:\mathrm{in}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]\:\mathrm{is} \\ $$ Answered by mrW1 last updated on 01/Jul/17 $${n}\mid\mathrm{sin}\:{x}\mid\:=\:{m}\mid\mathrm{sin}\:{x}\mid \\…
Question Number 17152 by Tinkutara last updated on 01/Jul/17 $$\mathrm{If}\:\mathrm{sin}{A}\:=\:\mathrm{sin}{B}\:\mathrm{and}\:\mathrm{cos}{A}\:=\:\mathrm{cos}{B},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{A}\:=\:{B}\:+\:{n}\pi,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{2}\right)\:{A}\:=\:{B}\:−\:{n}\pi,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{3}\right)\:{A}\:=\:\mathrm{2}{n}\pi\:+\:{B},\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{4}\right)\:{A}\:=\:{n}\pi\:−\:{B},\:{n}\:\in\:{I} \\ $$ Answered by ajfour last updated…
Question Number 17150 by Tinkutara last updated on 01/Jul/17 $$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{cos}\:{x}\:+\:\mathrm{3}\:\mathrm{cos}\:\left(\mathrm{2}{Kx}\right)\:=\:\mathrm{4} \\ $$$$\mathrm{has}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{solution},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{K}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rational}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form} \\ $$$$\frac{{P}}{{P}\:+\:\mathrm{1}},\:{P}\:\neq\:−\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{K}\:\mathrm{is}\:\mathrm{irrational}\:\mathrm{number}\:\mathrm{whose} \\ $$$$\mathrm{rational}\:\mathrm{approximation}\:\mathrm{does}\:\mathrm{not} \\ $$$$\mathrm{exceed}\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:{K}\:\mathrm{is}\:\mathrm{irrational}\:\mathrm{number} \\…