Question Number 127787 by bemath last updated on 02/Jan/21 $$\:\mathrm{csc}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:=?\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127684 by liberty last updated on 01/Jan/21 $$\mathrm{P}\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{15}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{15}}\right)\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{15}}\right)\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{15}}\right)\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{15}}\right)\mathrm{cos}\:\left(\frac{\mathrm{6}\pi}{\mathrm{15}}\right)\mathrm{cos}\:\left(\frac{\mathrm{7}\pi}{\mathrm{15}}\right) \\ $$$$\mathrm{P}=?\: \\ $$ Answered by bramlexs22 last updated on 01/Jan/21 Commented by bobhans last…
Question Number 127542 by Study last updated on 30/Dec/20 Commented by Study last updated on 30/Dec/20 $${help}\:{me} \\ $$ Commented by Study last updated on…
Question Number 192985 by MM42 last updated on 01/Jun/23 $${prove}\:{it}\:: \\ $$$${lim}_{{n}\rightarrow\infty} \:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{\theta}{\mathrm{2}^{{i}} }=\frac{{sin}\theta}{\theta} \\ $$$${then}\:{show}\:: \\ $$$${im}_{{n}\rightarrow\infty} \:{cos}\frac{\pi}{\mathrm{4}}{cos}\frac{\pi}{\mathrm{8}}…{cos}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\:=\frac{\mathrm{2}}{\pi} \\ $$$$ \\…
Question Number 127402 by bemath last updated on 29/Dec/20 $$\:{Compute}\:{x}\:{if}\:\mathrm{arctan}\:{x}+\mathrm{arctan}\:\mathrm{1}\:=\:\mathrm{2}\left(\mathrm{arctan}\:{x}−\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right)\: \\ $$ Answered by liberty last updated on 29/Dec/20 $$\Leftrightarrow\:\mathrm{arctan}\:{x}+\mathrm{arctan}\:\mathrm{1}=\mathrm{2arctan}\:{x}−\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{arctan}\:{x}\:=\:\mathrm{arctan}\:\mathrm{1}+\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\left(\mathrm{arctan}\:{x}\right)\:=\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{2arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right) \\…
Question Number 127383 by Lordose last updated on 29/Dec/20 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{analytic}\:\mathrm{proof}\:\mathrm{of}\:\mathrm{the}\:\mathrm{result} \\ $$$$\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{y}\right)\:+\:\mathrm{sin}\left(\mathrm{y}\right)\mathrm{cos}\left(\mathrm{x}\right) \\ $$ Answered by bemath last updated on 29/Dec/20 Answered by liberty last…
Question Number 127373 by mnjuly1970 last updated on 29/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{elemeary}\:\:\:{calculus}.. \\ $$$$\:\:{if}\:\:\:\:\begin{cases}{{sin}\left(\mathrm{3}{x}\right)+{cos}\left(\mathrm{3}{x}\right)={m}}\\{{sin}\left({x}\right)+{cos}\left({x}\right)={n}}\end{cases} \\ $$$$\:\:\:{then}\:\:\:{find}\:{the}\:{relatiomship} \\ $$$$\:\:\:{between}\:\:'{m}'\:{and}\:'\:{n}'\:{independent}\:{of} \\ $$$$\:\:'\:{x}'\:…. \\ $$ Answered by bemath last updated…
Question Number 61835 by alphaprime last updated on 09/Jun/19 $$\mathrm{If}\:\alpha\:=\:\mathrm{Cis}\left(\mathrm{2}\pi/\mathrm{7}\right)\:\mathrm{and}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{A}_{\mathrm{0}} \:+\:\sum_{\mathrm{n}=\mathrm{1}} ^{\mathrm{14}} \mathrm{A}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \: \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}\:\sum_{\alpha=\mathrm{0}} ^{\mathrm{6}} \mathrm{f}\left(\alpha^{\mathrm{n}} \mathrm{x}\right)=\:\mathrm{7}\left(\mathrm{A}_{\mathrm{0}} +\mathrm{A}_{\mathrm{7}} \mathrm{x}^{\mathrm{7}} +\mathrm{A}_{\mathrm{14}} \mathrm{x}^{\mathrm{14}} \right)…
Question Number 61799 by alphaprime last updated on 08/Jun/19 $$\mathrm{If}\:\mathrm{a}\:=\:\mathrm{Cos}\alpha\:−\mathrm{iSin}\alpha\:\mathrm{and}\:\mathrm{b}\:=\:\mathrm{Cos}\beta\:−\mathrm{iSin}\beta \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\frac{\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{1}−\mathrm{ab}\right)}{\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{1}+\mathrm{ab}\right)}\:=\:\frac{\mathrm{Sin}\alpha+\mathrm{Sin}\beta}{\mathrm{Sin}\alpha−\mathrm{Sin}\beta} \\ $$ Answered by tanmay last updated on 09/Jun/19 $${a}={e}^{−{i}\alpha} \:\:\:{b}={e}^{−{i}\beta} \\ $$$$\frac{{a}}{{b}}={e}^{−{i}\left(\propto−\beta\right)}…
Question Number 127307 by Study last updated on 28/Dec/20 Commented by Dwaipayan Shikari last updated on 28/Dec/20 $${n}=\mathrm{23} \\ $$ Commented by Study last updated…