Question Number 130469 by liberty last updated on 26/Jan/21 $$\:{P}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$ Answered by EDWIN88 last updated on 26/Jan/21 $$\mathrm{2psin}\:\frac{\pi}{\mathrm{15}}=\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\pi}{\mathrm{3}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$$$\mathrm{4psin}\:\frac{\pi}{\mathrm{15}}=\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}.\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\ $$$$\mathrm{8psin}\:\frac{\pi}{\mathrm{15}}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{15}}.\mathrm{cos}\:\frac{\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}.\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}} \\…
Question Number 130423 by john_santu last updated on 25/Jan/21 $$\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}\sqrt{{x}}}{\mathrm{3}{x}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}}\:\right)=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$ Answered by liberty last updated on 25/Jan/21 $$\:{let}\:{p}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}\sqrt{{x}}}\right)\:;\:{q}=\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}}\:\right)…
Question Number 64773 by ankan0 last updated on 21/Jul/19 $$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}/\sqrt{\left.\mathrm{5}\right)}+\mathrm{cot}^{−\mathrm{1}} \mathrm{3}\right. \\ $$ Answered by Kunal12588 last updated on 21/Jul/19 $$\mathrm{cot}^{−\mathrm{1}} \left({a}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\right)…
Question Number 130221 by bramlexs22 last updated on 23/Jan/21 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{tan}\:\mathrm{2A}=\frac{\mathrm{2tan}\:\mathrm{A}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{A}} \\ $$ Answered by EDWIN88 last updated on 23/Jan/21 $$\mathrm{by}\:\mathrm{De}'\mathrm{Moivre}\:\mathrm{theorem}\: \\ $$$$\mathrm{cos}\:\mathrm{2A}+{i}\:\mathrm{sin}\:\mathrm{2}{A}\:=\:\left(\mathrm{cos}\:\mathrm{A}+{i}\:\mathrm{sin}\:\mathrm{A}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{cos}^{\mathrm{2}}…
Question Number 64619 by Aysha Abedin last updated on 19/Jul/19 $${show}\:{that}\:\int_{−\alpha} ^{\alpha} {sinc}\left({x}\right){dx}=\int_{−\alpha} ^{\alpha} {sinc}^{\mathrm{2}} \left({x}\right){dx}=\Pi \\ $$ Commented by mathmax by abdo last updated…
Question Number 130082 by benjo_mathlover last updated on 22/Jan/21 $$\mathrm{Given}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:{a}+\mathrm{sin}\:{b}\right)+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:{a}−\mathrm{sin}\:{b}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:^{\mathrm{2}} {a}\:+\mathrm{sin}\:^{\mathrm{2}} {b}. \\ $$ Answered by liberty last updated on 22/Jan/21…
Question Number 130054 by EDWIN88 last updated on 22/Jan/21 $$\mathrm{The}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{x}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{x}\right)\:\mathrm{is}\:? \\ $$ Answered by liberty last updated on 22/Jan/21 Commented by MJS_new last…
Question Number 130047 by liberty last updated on 22/Jan/21 $$\:\mathrm{Given}\:\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}\:\mathrm{x}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{3sin}\:\mathrm{x} \\ $$$$\mathrm{where}\:−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{all}\:\mathrm{posibble}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$ Answered by MJS_new last updated on 22/Jan/21…
Question Number 130041 by liberty last updated on 22/Jan/21 $$\mathrm{Given}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{of}\:\mathrm{equation}\:{a}\:\mathrm{tan}\:{x}\:+\:{b}\:\mathrm{sec}\:{x}\:=\:{c}\: \\ $$$${find}\:\mathrm{tan}\:\left({A}+\mathrm{B}\right). \\ $$ Answered by nueron last updated on 22/Jan/21 Commented by…
Question Number 64498 by slotto last updated on 18/Jul/19 $$\mathrm{please},\:\mathrm{anyone}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\mathrm{i}.\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{cos}\left(\frac{\mathrm{k}^{\mathrm{2}} \pi}{\mathrm{n}}\right) \\ $$$$\mathrm{ii}.\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\Sigma}}\mathrm{sin}\left(\frac{\mathrm{k}^{\mathrm{2}} \pi}{\mathrm{n}}\right) \\ $$$$\mathrm{thank}\:\mathrm{you}. \\ $$ Terms…