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) \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{1}+\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}}…
Question Number 188589 by Rupesh123 last updated on 03/Mar/23 Commented by universe last updated on 03/Mar/23 $${Question}\:{no}\:\mathrm{188339} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 57474 by malwaan last updated on 05/Apr/19 $$\mathrm{prove}\: \\ $$$$\boldsymbol{{sin}}\mathrm{18}×\boldsymbol{{cos}}\mathrm{36}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by Abdo msup. last updated on 05/Apr/19 $$\mathrm{18}\rightarrow{x} \\ $$$$\mathrm{180}\rightarrow\pi\:\Rightarrow\mathrm{180}{x}\:=\mathrm{18}\pi\:\Rightarrow{x}\:=\frac{\pi}{\mathrm{10}}\:\Rightarrow\mathrm{36}\:=\frac{\pi}{\mathrm{5}}…
Question Number 57469 by olalekan2 last updated on 05/Apr/19 $$\frac{\mathrm{4}{tan}\mathrm{75}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{75}}=\frac{\mathrm{1}}{{cos}\mathrm{150}}\: \\ $$$${find}\:{tan}\mathrm{75}\:{in}\:{surd}\:{form} \\ $$ Commented by kaivan.ahmadi last updated on 05/Apr/19 $${if}\:{you}\:{want}\:{to}\:{find}\:{tv}\mathrm{75} \\ $$$${you}\:{can}\:{solve}:…
Question Number 188475 by cortano12 last updated on 02/Mar/23 $$\:\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{4x}+\sqrt{\mathrm{x}}\:\right)\right)\mathrm{cos}\:\left(\pi\left(\mathrm{x}+\mathrm{7}\sqrt{\mathrm{x}}\right)\right)=\mathrm{1} \\ $$$$\:\mathrm{x}=? \\ $$ Answered by Frix last updated on 02/Mar/23 $$\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:=\mathrm{1} \\ $$$$\Leftrightarrow \\…
Question Number 57383 by Tinkutara last updated on 03/Apr/19 Commented by Tinkutara last updated on 03/Apr/19 B part Answered by tanmay.chaudhury50@gmail.com last updated on 03/Apr/19 $${cos}^{−\mathrm{1}}…