Question Number 143063 by ERA last updated on 09/Jun/21 $$\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)…..\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{n}\pi}}{\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{\mathrm{n}}} } \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143057 by ERA last updated on 09/Jun/21 $$\mathrm{cos}\left(\boldsymbol{\alpha}\right)×\mathrm{cos}\left(\mathrm{2}\alpha\right)×\mathrm{cos}\left(\mathrm{4}\alpha\right)×….×\mathrm{cos}\left(\mathrm{2}^{\mathrm{n}} \boldsymbol{\alpha}\right)=\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \boldsymbol{\alpha}\right)}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \mathrm{sin}\left(\alpha\right)} \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$ Answered by Dwaipayan Shikari last updated on 09/Jun/21…
Question Number 143039 by bramlexs22 last updated on 09/Jun/21 $$\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${x}=? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 77472 by mathocean1 last updated on 06/Jan/20 $$\mathrm{Hello}\:\mathrm{sirs}…\: \\ $$$$\mathrm{i}\:\mathrm{want}\:\mathrm{that}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{me}\:\mathrm{how}\: \\ $$$$\:\mathrm{we}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{trigonometric} \\ $$$$\mathrm{inequality}\:\mathrm{with}\:\mathrm{tangente}.\: \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{take}\:\mathrm{for}\:\mathrm{example}\:\mathrm{tan2x}\geqslant\sqrt{\mathrm{3}} \\ $$$$\mathrm{i}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{type}\:\mathrm{of}\:\mathrm{equality} \\ $$$$\mathrm{but}\:\mathrm{not}\:\mathrm{the}\:\mathrm{inequality}!\:\mathrm{Please}\: \\ $$$$\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help}…\:\mathrm{Even}\:\mathrm{the}\:\mathrm{steps} \\…
Question Number 143003 by bramlexs22 last updated on 08/Jun/21 Answered by EDWIN88 last updated on 08/Jun/21 $$\left(\mathrm{1}\right)\:\mathrm{Since}\:\Delta\mathrm{ACE}\:\mathrm{and}\:\Delta\mathrm{ACB}\:\mathrm{share}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{altitude},\mathrm{and}\:\mathrm{AE}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{AB}\:,\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{ACE}= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{ACB}\:.\: \\ $$$$\mathrm{By}\:\mathrm{Heron}\:\mathrm{formula}\: \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{ACB}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\mathrm{15}}{\mathrm{2}}\left(\frac{\mathrm{7}}{\mathrm{2}}\right)\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\frac{\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{4}}…
Question Number 11900 by @ANTARES_VY last updated on 04/Apr/17 $$\boldsymbol{\mathrm{Calculate}}. \\ $$$$\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\pi}}{\mathrm{7}}×\boldsymbol{\mathrm{cos}}\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{7}}×\boldsymbol{\mathrm{cos}}\frac{\mathrm{5}\boldsymbol{\pi}}{\mathrm{7}}. \\ $$ Answered by ajfour last updated on 04/Apr/17 $$=\:\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\left[−\mathrm{cos}\:\left(\pi−\frac{\mathrm{5}\pi}{\mathrm{7}}\right)\:\right] \\ $$$$=\:−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}} \\…
Question Number 11844 by minakshidahaval0202@gmail.co. last updated on 02/Apr/17 $$\mathrm{cot}\alpha+\mathrm{cosec}\alpha={k}\:{then}\:{find}\:\mathrm{cosec}\alpha−{cot}\alpha\:{and}\:{also}\:{find}\:{cot}\alpha \\ $$ Answered by sma3l2996 last updated on 02/Apr/17 $${we}\:{have} \\ $$$$\left({i}\right):{cot}\alpha+{cosec}\alpha=\frac{{cos}\alpha}{{sin}\alpha}+\frac{\mathrm{1}}{{sin}\alpha}={k} \\ $$$$=\frac{{cos}\alpha+\mathrm{1}}{{sin}\alpha}=\frac{\left({cos}\alpha+\mathrm{1}\right)\left({cos}\alpha−\mathrm{1}\right)}{{sin}\alpha\left({cos}\alpha−\mathrm{1}\right)}=\frac{−{sin}^{\mathrm{2}} \alpha}{{sin}\alpha\left({cos}\alpha−\mathrm{1}\right)}…
Question Number 11751 by tawa last updated on 31/Mar/17 $$\mathrm{Simplify}:\:\mathrm{sin}\left(\mathrm{6}\theta\right) \\ $$ Answered by mrW1 last updated on 31/Mar/17 $$\mathrm{sin}\:\left(\mathrm{6}\theta\right)=\mathrm{sin}\:\left(\mathrm{4}\theta+\mathrm{2}\theta\right)=\mathrm{sin}\:\left(\mathrm{4}\theta\right)\mathrm{cos}\:\left(\mathrm{2}\theta\right)+\mathrm{sin}\:\left(\mathrm{2}\theta\right)\mathrm{cos}\:\left(\mathrm{4}\theta\right) \\ $$$$=\mathrm{2sin}\:\left(\mathrm{2}\theta\right)\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)+\mathrm{sin}\:\left(\mathrm{2}\theta\right)\left[\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\right] \\…
Question Number 142794 by EDWIN88 last updated on 05/Jun/21 $$\mathrm{If}\:\mathrm{2cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0}\:\mathrm{and}\: \\ $$$$\mathrm{sin}\:\left(\mathrm{2x}−\mathrm{2y}\right)=\mathrm{cos}\:\mathrm{y}\:\mathrm{where}\:\frac{\pi}{\mathrm{4}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}\:\mathrm{and} \\ $$$$\frac{\pi}{\mathrm{4}}\leqslant\mathrm{y}\leqslant\frac{\pi}{\mathrm{2}}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{2x}+\mathrm{y}\right)}\\{\mathrm{cos}\:\left(\mathrm{2x}+\mathrm{y}\right)}\\{\mathrm{cos}\:\left(\mathrm{2x}−\mathrm{y}\right)}\\{\mathrm{sin}\:\left(\mathrm{2x}−\mathrm{y}\right)}\end{cases} \\ $$ Answered by Rasheed.Sindhi last updated on 05/Jun/21 $$\mathrm{2cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\…
Question Number 77218 by john santu last updated on 04/Jan/20 $${what}\:{is}\:{solution}\:{in}\:\mathbb{R}\: \\ $$$$\frac{\mathrm{sin}\:\left({x}\right)−\mid{x}+\mathrm{2}\mid}{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}}\geqslant\mathrm{0}\:?\: \\ $$ Answered by MJS last updated on 04/Jan/20 $$\mathrm{sin}\:{x}\:−\mid{x}+\mathrm{2}\mid\geqslant\mathrm{0}\wedge\left({x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}\right)>\mathrm{0}…