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Question Number 82277 by M±th+et£s last updated on 19/Feb/20 Commented by M±th+et£s last updated on 19/Feb/20 $${find}\:{the}\:{area}\:{of}\:\:{the}\:{rectangle} \\ $$ Commented by mr W last updated…
Question Number 16740 by Tinkutara last updated on 26/Jun/17 $$\mathrm{The}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{cos}\:\left(\mathrm{33}\pi\:+\:\theta\right)\right)\:+\:\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:\left(\mathrm{45}\pi\:+\:\theta\right)\right) \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{1} \\…
Question Number 147809 by mathdanisur last updated on 23/Jul/21 $$\frac{\boldsymbol{{sin}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:+\:\boldsymbol{{cos}}^{\mathrm{6}} \left(\boldsymbol{\alpha}\right)\:-\:\mathrm{1}}{\boldsymbol{{sin}}^{\mathrm{4}} \left(\boldsymbol{\alpha}\right)\:-\:\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{\alpha}\right)}\:=\:? \\ $$ Answered by gsk2684 last updated on 23/Jul/21 $$\left(\mathrm{sin}\:^{\mathrm{2}} \alpha\right)^{\mathrm{3}}…
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Question Number 16739 by Tinkutara last updated on 26/Jun/17 $$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{and}\:\mathrm{let}\:{A}', \\ $$$${B}'\:\mathrm{and}\:{C}'\:\mathrm{be}\:\mathrm{its}\:\mathrm{projections}\:\mathrm{onto}\:\mathrm{the} \\ $$$$\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB},\:\mathrm{respectively}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{triangles}\:{MAC}',\:{MBA}'\:\mathrm{and} \\ $$$${MCB}'\:\mathrm{equals}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{trianges}\:{MAB}',\:{MBC}'\:\mathrm{and} \\…
Question Number 16738 by Tinkutara last updated on 26/Jun/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{segments}\:\mathrm{joining}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equiangular}\:\mathrm{hexagon}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147811 by mathdanisur last updated on 23/Jul/21 Answered by Olaf_Thorendsen last updated on 23/Jul/21 $$\mathrm{X}\:=\:\mathrm{sin}\left(\frac{\mathrm{arccos}{x}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:\mathrm{X}^{\mathrm{2}} \:=\:\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{arccos}{x}}{\mathrm{4}}\right) \\ $$$$\mathrm{X}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{2}×\frac{\mathrm{arccos}{x}}{\mathrm{4}}\right)\right) \\…
Question Number 16737 by Tinkutara last updated on 26/Jun/17 $$\mathrm{A}\:\mathrm{convex}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{given}\:\mathrm{in}\:\mathrm{which} \\ $$$$\mathrm{any}\:\mathrm{two}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{property}:\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{their}\:\mathrm{midpoints}\:\mathrm{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{times}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{lengths}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{hexagon}\:\mathrm{is}\:\mathrm{equiangular}. \\ $$ Terms of Service…