Question Number 68390 by TawaTawa last updated on 10/Sep/19 Answered by som(math1967) last updated on 10/Sep/19 $${tan}\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}}\:\right)+{tan}\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}}…
Question Number 133851 by physicstutes last updated on 24/Feb/21 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:{r}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{subtends}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\alpha\:\mathrm{at}\:\mathrm{its}\:\mathrm{centre}.\:\mathrm{A}\:\mathrm{chord}\:\mathrm{cuts}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{it}\:\mathrm{divides}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{into}\:\mathrm{two}\:\mathrm{segments}\:\mathrm{of}\:\mathrm{areas}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{1}:\mathrm{5}.\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\mathrm{sin}\:\alpha\:=\:\alpha−\frac{\pi}{\mathrm{3}} \\ $$ Terms of Service Privacy Policy…
Question Number 133782 by bramlexs22 last updated on 24/Feb/21 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{function}\:\mathrm{y}=\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:+\mathrm{cos}\:\mathrm{x}\:+\mathrm{3}\: \\ $$$$\mathrm{on}\:\mathrm{interval}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$ Answered by bobhans last updated on 24/Feb/21 $$\:{y}=\left(\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 68078 by anaplak last updated on 04/Sep/19 Answered by Smail last updated on 04/Sep/19 $${C}+{iS}=\mathrm{1}+{z}\left({cos}\theta+{isin}\theta\right)+\frac{{z}^{\mathrm{2}} }{\mathrm{2}!}\left({cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta\right)+… \\ $$$$=\mathrm{1}+{ze}^{{i}\theta} +\frac{\left({ze}^{{i}\theta} \right)^{\mathrm{2}} }{\mathrm{2}!}+…={e}^{{ze}^{{i}\theta} } ={e}^{{zcos}\theta}…
Question Number 68077 by anaplak last updated on 04/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133587 by bemath last updated on 23/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}\:\mathrm{12}°\:. \\ $$ Answered by Dwaipayan Shikari last updated on 23/Feb/21 $${sin}\frac{\pi}{\mathrm{60}}={sin}\frac{\pi}{\mathrm{10}}{cos}\frac{\pi}{\mathrm{12}}−{sin}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{10}} \\ $$$$=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}.\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}.\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left(\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)−\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}\right)…
Question Number 68022 by anaplak last updated on 03/Sep/19 Commented by anaplak last updated on 08/Sep/19 $${your}\:{given}\:{solution}\:{is}\:{not}\:{right}\:{according}\:{to}\:{my}\:{question}\:.. \\ $$ Answered by mr W last updated…
Question Number 68021 by anaplak last updated on 03/Sep/19 Commented by Tanmay chaudhury last updated on 03/Sep/19 Answered by Tanmay chaudhury last updated on 03/Sep/19…
Question Number 133444 by mnjuly1970 last updated on 22/Feb/21 Answered by guyyy last updated on 22/Feb/21 Commented by mnjuly1970 last updated on 23/Feb/21 $${thank}\:{you}\:{so}\:{much}… \\…
Question Number 133443 by mnjuly1970 last updated on 22/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:\:\:\:\:{calculus}……. \\ $$$$\:\:{if}\:\:{a},{b},{c}\:\geqslant\mathrm{0} \\ $$$$\:\:\:\:{and}\:::\:\:\:\:{acos}^{\mathrm{2}} \left({x}\right)+{bsin}^{\mathrm{2}} \left({x}\right)\leqslant{c} \\ $$$$\:\:\:\:{then}\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\sqrt{{a}}\:{cos}^{\mathrm{2}} \left({x}\right)+\sqrt{{b}}\:{sin}^{\mathrm{2}} \left({x}\right)\leqslant\sqrt{{c}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………. \\…