Question Number 65345 by KanhAshish last updated on 28/Jul/19 $$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}^{\mathrm{2018}} \mathrm{x}\:+\mathrm{cos}^{\mathrm{2019}} \mathrm{x}\:? \\ $$ Commented by mr W last updated on 28/Jul/19 $${maximum}\:{of}\:\mathrm{sin}^{{m}} \:{x}+\mathrm{cos}^{{n}} \:{x}\:{is}\:\mathrm{1}.…
Question Number 130819 by EDWIN88 last updated on 29/Jan/21 $$\:\mathrm{8sin}\:^{\mathrm{3}} \left({x}+\frac{\pi}{\mathrm{6}}\right)=\:\mathrm{cos}\:\left(\mathrm{3}{x}\right) \\ $$$$\:{x}=? \\ $$ Answered by mr W last updated on 29/Jan/21 $${let}\:{u}={x}+\frac{\pi}{\mathrm{6}} \\…
Question Number 130632 by bramlexs22 last updated on 27/Jan/21 Answered by EDWIN88 last updated on 27/Jan/21 $$\mathrm{cos}\:\mathrm{4095}°\:=\:\mathrm{cos}\:\left(\mathrm{360}°×\mathrm{11}+\mathrm{135}°\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${then}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{4095}°\right)=\mathrm{sin}\:\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$\:=−\frac{\pi}{\mathrm{4}} \\…
Question Number 130589 by bramlexs22 last updated on 27/Jan/21 $$\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{x}\right)=\pi/\mathrm{2} \\ $$ Answered by EDWIN88 last updated on 27/Jan/21 $$\:{If}\:{f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \left({x}\right)+\mathrm{cot}^{−\mathrm{1}} \left({x}\right) \\…
Question Number 130581 by bramlexs22 last updated on 27/Jan/21 $$\mathrm{If}\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} \:,\mathrm{a}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}\:\mathrm{and}\:\mathrm{d}\:\mathrm{is}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{then}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{d}}{\mathrm{1}+\mathrm{a}_{\mathrm{1}} \mathrm{a}_{\mathrm{2}} }\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{d}}{\mathrm{1}+\mathrm{a}_{\mathrm{2}} \mathrm{a}_{\mathrm{3}} }\right)=? \\ $$ Answered by…
Question Number 130580 by bramlexs22 last updated on 27/Jan/21 $$\mathrm{If}\:\mathrm{x}\epsilon\:\left[−\mathrm{1},\:\mathrm{0}\right)\:\mathrm{then}\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2sin}^{−\mathrm{1}} \left(\mathrm{x}\right)=? \\ $$ Commented by EDWIN88 last updated on 27/Jan/21 $$\pi \\ $$…
Question Number 130469 by liberty last updated on 26/Jan/21 $$\:{P}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$ Answered by EDWIN88 last updated on 26/Jan/21 $$\mathrm{2psin}\:\frac{\pi}{\mathrm{15}}=\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\pi}{\mathrm{3}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$$$\mathrm{4psin}\:\frac{\pi}{\mathrm{15}}=\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$$$\mathrm{8psin}\:\frac{\pi}{\mathrm{15}}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\…
Question Number 130423 by john_santu last updated on 25/Jan/21 $$\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}\sqrt{{x}}}{\mathrm{3}{x}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}}\:\right)=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$ Answered by liberty last updated on 25/Jan/21 $$\:{let}\:{p}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}\sqrt{{x}}}\right)\:;\:{q}=\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}}\:\right)…
Question Number 64773 by ankan0 last updated on 21/Jul/19 $$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}/\sqrt{\left.\mathrm{5}\right)}+\mathrm{cot}^{−\mathrm{1}} \mathrm{3}\right. \\ $$ Answered by Kunal12588 last updated on 21/Jul/19 $$\mathrm{cot}^{−\mathrm{1}} \left({a}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\right)…
Question Number 130221 by bramlexs22 last updated on 23/Jan/21 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{tan}\:\mathrm{2A}=\frac{\mathrm{2tan}\:\mathrm{A}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{A}} \\ $$ Answered by EDWIN88 last updated on 23/Jan/21 $$\mathrm{by}\:\mathrm{De}'\mathrm{Moivre}\:\mathrm{theorem}\: \\ $$$$\mathrm{cos}\:\mathrm{2A}+{i}\:\mathrm{sin}\:\mathrm{2}{A}\:=\:\left(\mathrm{cos}\:\mathrm{A}+{i}\:\mathrm{sin}\:\mathrm{A}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{cos}^{\mathrm{2}}…