Question Number 57959 by rahul 19 last updated on 15/Apr/19 Answered by MJS last updated on 16/Apr/19 $$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:={y}\:\Rightarrow\:{x}=\mathrm{sin}\:\frac{{y}}{\mathrm{3}} \\ $$$$\mathrm{with}\:{x}\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\wedge\:{y}\in\left[−\frac{\mathrm{3}\pi}{\mathrm{2}};\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:\mathrm{increases}\:\mathrm{for}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\…
Question Number 189026 by Rupesh123 last updated on 10/Mar/23 Answered by Ar Brandon last updated on 10/Mar/23 $$\mathrm{cot}^{\mathrm{2}} {x}+\mathrm{1}=\mathrm{cosec}^{\mathrm{2}} {x}\: \\ $$$$\mid\mathrm{4}−{a}\mid+\mathrm{1}=\frac{\mid{a}\mid}{\mathrm{3}}\:,\:\mathrm{0}<{a}\leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}−{a}+\mathrm{1}=\frac{{a}}{\mathrm{3}}\:\Rightarrow{a}=\frac{\mathrm{15}}{\mathrm{4}} \\…
Question Number 189012 by TUN last updated on 10/Mar/23 $${Triangle}\:{ABC}\:{have}:\: \\ $$$${sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}=\sqrt{\mathrm{3}}\left({cosA}+{cosB}+{cosC}\right) \\ $$$$=>\:{Prove}\:{that}\:{ABC}\:{is}\:{equilateral}\:{triangle} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 57915 by Tinkutara last updated on 14/Apr/19 Answered by mr W last updated on 14/Apr/19 $$\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)+\mathrm{2}^{\mathrm{tan}^{\mathrm{2}} \:\theta} =\mathrm{0} \\ $$$${let}\:{t}=\mathrm{tan}^{\mathrm{2}} \:\theta\geqslant\mathrm{0}…
Question Number 188912 by Mingma last updated on 08/Mar/23 Answered by a.lgnaoui last updated on 09/Mar/23 $${posons}\:\:\frac{\theta}{\mathrm{3}}={x} \\ $$$$\mathrm{sin}\:\left(\frac{\pi−\theta}{\mathrm{3}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{sin}\:\left(\frac{\pi+\theta}{\mathrm{3}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right) \\ $$$$\:\: \\ $$$$\mathrm{sin}^{\mathrm{3}}…
Question Number 57664 by necx1 last updated on 09/Apr/19 $${A}\:{luminous}\:{point}\:{P}\:\:{is}\:{inside}\:{a}\:{circle}. \\ $$$${A}\:{ray}\:{emanates}\:{from}\:{P}\:{and}\:{after}\:{two} \\ $$$${reflections}\:{by}\:{the}\:{circle},{returns}\:{to}\:{P}. \\ $$$${If}\:\theta\:{be}\:{the}\:{angle}\:{of}\:{incidence},\:{a}=\:{the} \\ $$$${distance}\:{of}\:{P}\:{from}\:{the}\:{centre}\:{of}\:{the} \\ $$$${circle}\:{and}\:{b}={the}\:{distance}\:{of}\:{the}\:{centre} \\ $$$${from}\:{the}\:{point}\:{where}\:{the}\:{ray}\:{in}\:{its} \\ $$$${course}\:{crosses}\:{its}\:{diameter}\:{through}\:{P}. \\…
Question Number 188720 by Rupesh123 last updated on 06/Mar/23 Commented by mr W last updated on 06/Mar/23 $${i}\:{think}\:{it}'{s}\:{impossible}. \\ $$ Commented by ajfour last updated…
Question Number 188721 by Rupesh123 last updated on 06/Mar/23 Answered by Frix last updated on 06/Mar/23 $${x}\geqslant\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{0}\leqslant\sqrt{{x}}<\infty \\ $$$$−\mathrm{2}.\mathrm{73581}…\leqslant\mathrm{sin}\:{x}\:+\mathrm{2sin}\:\mathrm{2}{x}\:\leqslant\mathrm{2}.\mathrm{73581}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{Exact}\:\mathrm{values}\:\pm\frac{\left.\sqrt{\mathrm{6}\left(\mathrm{789}+\mathrm{43}\sqrt{\mathrm{129}}\right.}\right)}{\mathrm{32}}\right]…
Question Number 123160 by liberty last updated on 23/Nov/20 $$\:{Given}\:\mathrm{sin}\:\rho\:+\:\mathrm{cos}\:\rho\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:,\:{where}\: \\ $$$$\:\mathrm{0}\:<\:\rho\:<\:\frac{\pi}{\mathrm{4}}.\:{Find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\rho−\mathrm{sin}\:\rho.\: \\ $$ Answered by benjo_mathlover last updated on 23/Nov/20 $$\Rightarrow\left(\mathrm{sin}\:\rho+\mathrm{cos}\:\rho\right)^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2sin}\:\rho\mathrm{cos}\:\rho=\frac{\mathrm{16}}{\mathrm{9}}…
Question Number 188645 by Rupesh123 last updated on 04/Mar/23 Answered by witcher3 last updated on 05/Mar/23 $$\mathrm{let}\:\mathrm{a}=\mathrm{cos}\left(\frac{\pi}{\mathrm{7}}\right),\mathrm{b}=−\mathrm{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right);\mathrm{c}=\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right) \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{1}+\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}}…