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Author: Tinku Tara

1-1-pi-2-1-pi-1-2-1-pi-2-2-1-1-1-pi-1-2-1-pi-2-2-1-1-pi-Prove-or-disprove-

Question Number 137347 by Dwaipayan Shikari last updated on 01/Apr/21 $$\frac{\mathrm{1}}{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+…}}}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+…}}}=\frac{\mathrm{1}}{\pi}\:\: \\ $$$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$ Answered by MJS_new last updated…

Question-71809

Question Number 71809 by TawaTawa last updated on 20/Oct/19 Answered by tw000001 last updated on 21/Oct/19 $$\mathrm{I}\:\mathrm{have}\:\mathrm{showed}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{on}\:\mathrm{No}.\mathrm{71776}. \\ $$ Terms of Service Privacy Policy Contact:…

Question-71806

Question Number 71806 by ahmadshahhimat775@gmail.com last updated on 20/Oct/19 Answered by MJS last updated on 20/Oct/19 $$\left(−\mathrm{2}\right)^{\sqrt{\mathrm{2}}} =\left(\mathrm{2e}^{\mathrm{i}\pi} \right)^{\sqrt{\mathrm{2}}} =\mathrm{2}^{\sqrt{\mathrm{2}}} \mathrm{e}^{\mathrm{i}\pi\sqrt{\mathrm{2}}} = \\ $$$$=\mathrm{2}^{\sqrt{\mathrm{2}}} \mathrm{cos}\:\pi\sqrt{\mathrm{2}}\:+\mathrm{2}^{\sqrt{\mathrm{2}}}…

Given-cos-8-6cos-6-13cos-4-8cos-2-cos-7-5cos-5-8cos-3-1-2-then-what-the-value-of-tan-2-

Question Number 137338 by liberty last updated on 01/Apr/21 $$\mathrm{Given}\:\frac{\mathrm{cos}\:\mathrm{8}\theta+\mathrm{6cos}\:\mathrm{6}\theta+\mathrm{13cos}\:\mathrm{4}\theta+\mathrm{8cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\mathrm{7}\theta+\mathrm{5cos}\:\mathrm{5}\theta+\mathrm{8cos}\:\mathrm{3}\theta}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{what}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\mathrm{2}\theta\:? \\ $$ Answered by EDWIN88 last updated on 01/Apr/21 $$\mathrm{consider}\:\mathrm{numerator} \\ $$$$\Leftrightarrow\:\mathrm{cos}\:\mathrm{8}\theta+\mathrm{cos}\:\mathrm{6}\theta+\mathrm{5}\left(\mathrm{cos}\:\mathrm{6}\theta+\mathrm{cos}\:\mathrm{4}\theta\right)+\mathrm{8}\left(\mathrm{cos}\:\mathrm{4}\theta+\mathrm{cos}\:\mathrm{2}\theta\right) \\…