Question Number 9370 by tawakalitu last updated on 03/Dec/16 Answered by geovane10math last updated on 03/Dec/16 $$\mathrm{There}\:\mathrm{are}\:\mathrm{infinite}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sizes}\:\mathrm{10} \\ $$$$\mathrm{and}\:\mathrm{15}.\:\mathrm{It}\:\mathrm{depends}\:\mathrm{of}\:\mathrm{inclination}\:\mathrm{of}\:{X}\hat {{Z}M}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{w}\:\mathrm{depends}\:\mathrm{of}\: \\ $$$${X}\hat {{Z}M}.…
Question Number 74853 by Raxreedoroid last updated on 01/Dec/19 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{combine}\:\mathrm{2}{cos}\left(\mathrm{90}°{x}\right)+{cos}\left(\mathrm{180}°{x}\right) \\ $$$$\mathrm{into}\:\mathrm{a}\:\mathrm{form}\:\mathrm{of}\:\:{a}\centerdot{cos}\left({b}\centerdot{x}+{c}\right) \\ $$$$\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}{x}\right)+{cos}\left(\pi{x}\right)\overset{?} {=}{a}\centerdot{cos}\left({bx}+{c}\right) \\ $$ Commented by mr W last updated on 01/Dec/19…
Question Number 140382 by john_santu last updated on 07/May/21 $$\mathrm{sin}\:^{\mathrm{10}} {x}\:+\:\mathrm{cos}\:^{\mathrm{10}} {x}\:=\:\frac{\mathrm{61}}{\mathrm{256}} \\ $$ Answered by MJS_new last updated on 07/May/21 $$\mathrm{sin}\:{x}\:={s}\wedge\mathrm{cos}\:{x}\:=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} } \\ $$$$\mathrm{5}{s}^{\mathrm{8}}…
Question Number 140378 by benjo_mathlover last updated on 07/May/21 $$\sqrt{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}}−\sqrt{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{for}\:\mathrm{x}\in\:\left[\:\mathrm{0},\mathrm{2}\pi\:\right] \\ $$ Commented by john_santu last updated on 07/May/21…
Question Number 140368 by meetbhavsar25 last updated on 06/May/21 Answered by benjo_mathlover last updated on 07/May/21 $$\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\right)\:=\:\mathrm{u} \\ $$$$\Rightarrow\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\:=\:\mathrm{cos}\:\mathrm{u} \\ $$$$\Rightarrow\mathrm{3}+\mathrm{5cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}+\mathrm{3cos}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\Rightarrow\left(\mathrm{5}−\mathrm{3cos}\:\mathrm{u}\right)\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}−\mathrm{3} \\…
Question Number 140338 by Arzoon last updated on 06/May/21 $${if}\:\theta=\frac{\pi}{\mathrm{2}{n}+\mathrm{1}},\:{n}\geqslant\mathrm{1},\:{n}\in{N}\:,{then}\:{prove}\:{that}: \\ $$$${tan}\theta{tan}\mathrm{2}\theta{tan}\mathrm{3}\theta\centerdot\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:{tan}\left({n}\theta\right)\:=\:\sqrt{\mathrm{2}{n}+\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9211 by tawakalitu last updated on 23/Nov/16 $$\mathrm{If} \\ $$$$\mathrm{sinA}\:+\:\mathrm{sinB}\:=\:\mathrm{a} \\ $$$$\mathrm{tanA}\:+\:\mathrm{tanB}\:=\:\mathrm{b} \\ $$$$\mathrm{secA}\:+\:\mathrm{secB}\:=\:\mathrm{c} \\ $$$$\mathrm{Prove}\:\mathrm{that}, \\ $$$$\mathrm{8bc}\:=\:\mathrm{a}\left[\mathrm{4b}^{\mathrm{2}} \:+\:\left(\mathrm{b}^{\mathrm{2}} \:−\:\mathrm{c}^{\mathrm{2}} \right)\right]^{\mathrm{2}} \\ $$…
Question Number 140192 by meetbhavsar25 last updated on 05/May/21 Answered by mr W last updated on 05/May/21 $${say}\:\mathrm{tan}^{−\mathrm{1}} {x}={t} \\ $$$${x}=\mathrm{tan}\:{t} \\ $$$$\mathrm{cos}\:{t}=\mathrm{tan}\:{t}=\frac{\mathrm{sin}\:{t}}{\mathrm{cos}\:{t}} \\ $$$$\mathrm{cos}^{\mathrm{2}}…
Question Number 140177 by liberty last updated on 05/May/21 $$\:\:\:\mathrm{solution}\:\mathrm{set}\:\mathrm{equation} \\ $$$$\:\:\:\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$ Answered by som(math1967) last updated on 05/May/21 $$\left({sin}^{\mathrm{2}}…
Question Number 9085 by tawakalitu last updated on 17/Nov/16 Answered by mrW last updated on 17/Nov/16 $${x}^{\mathrm{2}} +\left(\mathrm{2}\frac{\mathrm{1}}{\mathrm{8}}−{x}\right)^{\mathrm{2}} =\left(\mathrm{1}\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left(\frac{\mathrm{17}}{\mathrm{8}}−{x}\right)^{\mathrm{2}} −\left(\frac{\mathrm{13}}{\mathrm{8}}\right)^{\mathrm{2}} =\mathrm{0} \\…