Question Number 204303 by MASANJAJJ last updated on 11/Feb/24 Answered by SWPlaysMC last updated on 12/Feb/24 $$\mathrm{Duplicate}\:\mathrm{of}\:\mathrm{Q204302} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204276 by MASANJAJJ last updated on 11/Feb/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 204293 by depressiveshrek last updated on 11/Feb/24 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{trig}\:\mathrm{identity}: \\ $$$$\frac{\mathrm{2sin}\alpha+\mathrm{sin3}\alpha+\mathrm{sin5}\alpha}{\mathrm{cos}\alpha−\mathrm{2cos2}\alpha+\mathrm{cos3}\alpha}=\frac{\mathrm{2cos2}\alpha}{\mathrm{tan}\frac{\alpha}{\mathrm{2}}} \\ $$ Commented by Frix last updated on 11/Feb/24 $$\mathrm{Try}. \\ $$$$\alpha=\frac{\pi}{\mathrm{6}}\:\Rightarrow\:\mathrm{lhs}=−\mathrm{5}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:\:\:\:\:\mathrm{rhs}=\mathrm{2}+\sqrt{\mathrm{3}} \\…
Question Number 204278 by liuxinnan last updated on 11/Feb/24 $$\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{3}{x}+\mathrm{cos}\:\mathrm{5}{x}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\mathrm{sin}\:{x}+\mathrm{sin3}{x}+\:\mathrm{sin}\:\mathrm{5}{x}=\mathrm{1} \\ $$$$\mathrm{tan}\:\mathrm{3}{x}=? \\ $$ Commented by mr W last updated on 11/Feb/24 $${eq}.\left({i}\right)\:{and}\:{eq}.\left({ii}\right)\:{are}\:{not}\:{consistent}.…
Question Number 204279 by liuxinnan last updated on 11/Feb/24 Commented by liuxinnan last updated on 11/Feb/24 $${a}\:{beautiful}\:{image} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204275 by Frix last updated on 11/Feb/24 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\sqrt{\mathrm{tan}\:{x}}\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}\:{dx}=\left(\frac{\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\right)\pi \\ $$ Answered by witcher3 last updated on 11/Feb/24 $$\mathrm{nice}\:\mathrm{problem} \\…
Question Number 204270 by universe last updated on 10/Feb/24 Answered by witcher3 last updated on 10/Feb/24 $$\left(\mathrm{1}\right) \\ $$$$\mathrm{withe}\:\mathrm{recursion}\:\mathrm{We}\:\mathrm{can}\:\mathrm{easly}\:\mathrm{proov}\:\mathrm{that}\:\mathrm{a}_{\mathrm{n}} >\mathrm{0} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\mathrm{f}\left(\mathrm{a}_{\mathrm{n}} \right);\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:\mathrm{increase}\:\mathrm{function} \\…
Question Number 204264 by SANOGO last updated on 10/Feb/24 $$\underset{} {\int}{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$ Answered by witcher3 last updated on 10/Feb/24 $$\mathrm{by}\:\mathrm{part}\:\mathrm{u}=\mathrm{ln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right);\mathrm{v}'\left(\mathrm{x}\right)=\mathrm{1} \\ $$$$=\mathrm{xln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}}…
Question Number 204249 by universe last updated on 10/Feb/24 Commented by Frix last updated on 10/Feb/24 $${P}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{7}} }{\mathrm{16}}−\frac{\mathrm{21}{x}^{\mathrm{5}} }{\mathrm{16}}+\frac{\mathrm{35}{x}^{\mathrm{3}} }{\mathrm{16}}−\frac{\mathrm{35}{x}}{\mathrm{16}} \\ $$$${Q}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{16}}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{29}{x}}{\mathrm{16}}+\mathrm{1} \\…
Question Number 204250 by Noorzai last updated on 10/Feb/24 Answered by AST last updated on 10/Feb/24 $${tan}\left(\mathrm{9}\right)+{tan}\left(\mathrm{81}\right)−\left({tan}\mathrm{27}+{tan}\mathrm{63}\right) \\ $$$$=\frac{{sin}\mathrm{9}}{{cos}\mathrm{9}}+\frac{{sin}\mathrm{81}}{{cos}\mathrm{81}}−\left(\frac{{sin}\mathrm{27}}{{cos}\mathrm{27}}+\frac{{sin}\mathrm{63}}{{cos}\mathrm{63}}\right) \\ $$$$=\frac{\mathrm{2}{sin}\left(\mathrm{9}+\mathrm{81}\right)=\mathrm{2}}{\mathrm{2}{cos}\mathrm{9}{cos}\mathrm{81}}−\frac{\mathrm{2}{sin}\left(\mathrm{27}+\mathrm{63}\right)=\mathrm{2}}{\mathrm{2}{cos}\left(\mathrm{27}\right){cos}\left(\mathrm{63}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{9}\right){cos}\left(\mathrm{9}\right)}−\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{27}\right){cos}\left(\mathrm{27}\right)}=\frac{\mathrm{2}}{{sin}\left(\mathrm{18}\right)}−\frac{\mathrm{2}}{{sin}\left(\mathrm{54}\right)} \\ $$$${Let}\:\theta=\mathrm{18}°\Rightarrow{sin}\left(\mathrm{5}\theta\right)=\mathrm{16}{sin}^{\mathrm{5}}…