Question Number 18892 by Tinkutara last updated on 01/Aug/17 $$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{sin6}{x}\:+\:\mathrm{cos4}{x}\:=\:−\mathrm{2}\:\mathrm{have} \\ $$$$\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{nonnegative}\:\mathrm{solutions}\:{x}_{{k}} '\mathrm{s}, \\ $$$$\mathrm{where}\:\mathrm{0}\:\leqslant\:{x}_{\mathrm{1}} \:<\:{x}_{\mathrm{2}} \:<\:{x}_{\mathrm{3}} \:<\:….\:<\:{x}_{{k}} \:<\:{x}_{{k}+\mathrm{1}} \\ $$$$…..,\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\mathrm{1000}} {\sum}}\mid{x}_{{k}+\mathrm{1}} \:−\:{x}_{{k}} \mid\:\mathrm{is}…
Question Number 18847 by Tinkutara last updated on 30/Jul/17 $$\mathrm{In}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{usual} \\ $$$$\mathrm{notations}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{27}{r}^{\mathrm{2}} {R}}{{r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} }\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on 31/Jul/17…
Question Number 18841 by mondodotto@gmail.com last updated on 30/Jul/17 Answered by Tinkutara last updated on 31/Jul/17 $$\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{8}\theta\right)}}} \\ $$$$=\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{4}\theta\right)}}\:=\:\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{2}\theta\right)} \\ $$$$=\:\mathrm{2}\:\mathrm{cos}\:\theta \\ $$ Terms of…
Question Number 18830 by Joel577 last updated on 30/Jul/17 $$\mathrm{If}\:\mathrm{sin}\:\mathrm{2}{x}\:=\:{a} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{simplest}\:\mathrm{expression}\:\mathrm{for}\:\mathrm{cos}\:{x} \\ $$ Commented by mrW1 last updated on 30/Jul/17 $$\mathrm{cos}\:\mathrm{2x}=\pm\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{2cos}^{\mathrm{2}}…
Question Number 18797 by Tinkutara last updated on 29/Jul/17 $$\mathrm{If}\:{A},\:{B},\:{C}\:\mathrm{are}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}, \\ $$$$\mathrm{then}\:\mathrm{2sin}\frac{{A}}{\mathrm{2}}\mathrm{cosec}\frac{{B}}{\mathrm{2}}\mathrm{sin}\frac{{C}}{\mathrm{2}}\:−\:\mathrm{sin}{A}\mathrm{cot}\frac{{B}}{\mathrm{2}} \\ $$$$−\:\mathrm{cos}\:{A}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Independent}\:\mathrm{of}\:{A},\:{B},\:{C} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Function}\:\mathrm{of}\:{A},\:{B},\:{C} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Function}\:\mathrm{of}\:{A},\:{B} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Function}\:\mathrm{of}\:{B},\:{C} \\ $$ Answered…
Question Number 18742 by Tinkutara last updated on 31/Jul/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cot16}°\mathrm{cot44}°\:+\:\mathrm{cot44}°\mathrm{cot76}° \\ $$$$−\:\mathrm{cot76}°\mathrm{cot16}°\:\mathrm{is} \\ $$ Answered by Tinkutara last updated on 31/Jul/17 $$\frac{\mathrm{3}\:+\:\mathrm{cot}\:\mathrm{76}°\:\mathrm{cot}\:\mathrm{16}°}{\mathrm{cot}\:\mathrm{76}°\:+\:\mathrm{cot}\:\mathrm{16}°} \\ $$$$=\:\frac{\mathrm{3}\:\mathrm{sin}\:\mathrm{76}°\:\mathrm{sin}\:\mathrm{16}°\:+\:\mathrm{cos}\:\mathrm{76}°\:\mathrm{cos}\:\mathrm{16}°}{\mathrm{sin}\:\left(\mathrm{76}°\:+\:\mathrm{16}°\right)} \\…
Question Number 18723 by mondodotto@gmail.com last updated on 28/Jul/17 Answered by Tinkutara last updated on 29/Jul/17 $$\mathrm{2sin2}\theta\:+\:\mathrm{15cos2}\theta\:=\:\mathrm{10} \\ $$$$\frac{\mathrm{2sin2}\theta}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{15cos2}\theta}{\:\sqrt{\mathrm{229}}}\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}} \\ $$$$\mathrm{cos}\left(\mathrm{2}\theta\:−\:\alpha\right)\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}}\:,\:\mathrm{where}\:\alpha\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right) \\ $$$$\theta\:=\:{n}\pi\:\pm\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right)…
Question Number 18721 by Tinkutara last updated on 28/Jul/17 $$\mathrm{Let}\:{ABC}\:\mathrm{and}\:{ABC}'\:\mathrm{be}\:\mathrm{two}\:\mathrm{non}- \\ $$$$\mathrm{congruent}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides}\:{AB}\:=\:\mathrm{4}, \\ $$$${AC}\:=\:{AC}'\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{angle}\:{B}\:=\:\mathrm{30}°. \\ $$$$\mathrm{The}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{is} \\ $$ Answered by ajfour last updated…
Question Number 18719 by Tinkutara last updated on 28/Jul/17 $$\mathrm{If}\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{1}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{2}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{3}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{4}} \\ $$$$+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{5}} \:=\:\mathrm{5},\:\mathrm{then}\:\mathrm{sin}\:{x}_{\mathrm{1}} \:+\:\mathrm{2sin}\:{x}_{\mathrm{2}} \:+ \\ $$$$\mathrm{3sin}\:{x}_{\mathrm{3}}…
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