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P-1-1-i-P-n-1-P-n-1-P-n-1-lim-n-Im-P-n-

Question Number 59858 by jimful last updated on 15/May/19 $$\mathrm{P}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{{i}}\: \\ $$$$\mathrm{P}_{\mathrm{n}+\mathrm{1}} \mathrm{P}_{\mathrm{n}} =\mathrm{1}−\mathrm{P}_{\mathrm{n}+\mathrm{1}} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}Im}\left(\mathrm{P}_{\mathrm{n}} \right)\:=? \\ $$ Terms of Service Privacy Policy…

x-y-5pi-3-sin-x-2sin-y-

Question Number 125359 by weltr last updated on 10/Dec/20 $$\begin{cases}{{x}\:+\:{y}\:=\:\frac{\mathrm{5}\pi}{\mathrm{3}}}\\{{sin}\:{x}\:=\:\mathrm{2}{sin}\:{y}}\end{cases} \\ $$ Answered by Ar Brandon last updated on 10/Dec/20 $$\begin{cases}{\mathrm{x}+\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{3}}}&{…\left(\mathrm{i}\right)}\\{\mathrm{sinx}=\mathrm{2siny}}&{…\left(\mathrm{ii}\right)}\end{cases} \\ $$$$\mathrm{From}\:\left(\mathrm{i}\right),\:\mathrm{x}+\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{3}}\:\Rightarrow\:\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{3}}−\mathrm{x} \\ $$$$\mathrm{substituting}\:\mathrm{the}\:\mathrm{above}\:\mathrm{result}\:\mathrm{in}\:\left(\mathrm{ii}\right)\:\mathrm{we}\:\mathrm{have}\:;…

6-3-3-1-3-9-1-3-simplify-

Question Number 59812 by ANTARES VY last updated on 15/May/19 $$\frac{\mathrm{6}}{\mathrm{3}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}\:\:\:\boldsymbol{\mathrm{simplify}}. \\ $$ Commented by Kunal12588 last updated on 15/May/19 $${let}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}={a} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{9}}={a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}={a}^{\mathrm{3}}…

tg-2-590-cos-2-320-sin-111-cos-159-cos-279-sin-549-ctg-950-sin-2-400-simplify-

Question Number 59811 by ANTARES VY last updated on 15/May/19 $$\left(\frac{\boldsymbol{\mathrm{tg}}^{\mathrm{2}} \left(\mathrm{590}°\right)}{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\mathrm{320}°\right)}+\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{111}°\right)}{\boldsymbol{\mathrm{cos}}\left(\mathrm{159}°\right)}\right)\left(\frac{\boldsymbol{\mathrm{cos}}\left(\mathrm{279}°\right)}{\boldsymbol{\mathrm{sin}}\left(\mathrm{549}°\right)}+\frac{\boldsymbol{\mathrm{ctg}}\left(\mathrm{950}°\right)}{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\mathrm{400}°\right)}\right) \\ $$$$\boldsymbol{\mathrm{simplify}}. \\ $$ Commented by Kunal12588 last updated on 15/May/19…