Question Number 94523 by peter frank last updated on 19/May/20 Commented by PRITHWISH SEN 2 last updated on 19/May/20 $$\mathrm{By}\:\mathrm{Napier}'\mathrm{s}\:\mathrm{Analogy}\: \\ $$$$\mathrm{for}\:\mathrm{any}\bigtriangleup\mathrm{ABC} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{B}}−\boldsymbol{\mathrm{C}}}{\mathrm{2}}\:=\frac{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}\:\boldsymbol{\mathrm{cot}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}} \\…
Question Number 160050 by tounghoungko last updated on 24/Nov/21 Answered by mindispower last updated on 24/Nov/21 $${cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right) \\ $$$${sin}\left({a}\right)+\mathrm{sin}\left(\mathrm{b}\right)=\mathrm{2cos}\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right) \\ $$$$\Leftrightarrow\frac{{cos}\left(\mathrm{3}{x}\right)\left(\mathrm{2}{cos}\left(\mathrm{5}{x}\right)+\mathrm{1}\right)}{{cos}\left(\mathrm{3}{x}\right)\left(\mathrm{2}{sin}\left(\mathrm{5}{x}\right)+\mathrm{1}\right)}=\mathrm{1} \\ $$$$\forall{x}\in\mathbb{R}−\left\{{cos}\left(\mathrm{3}{x}\right)=\mathrm{0}\right\} \\ $$$$\Leftrightarrow{sin}\left(\mathrm{5}{x}\right)={cos}\left(\mathrm{5}{x}\right)\:{easy}\:{now}…
Question Number 159928 by abdullah_ff last updated on 22/Nov/21 $$ \\ $$$$ \\ $$ Answered by TheSupreme last updated on 22/Nov/21 $${we}\:{have}\:{info}\:{about}\:{angles} \\ $$$${there}\:{are}\:{infinite}\:{congruent}\:{triangles} \\…
Question Number 94344 by i jagooll last updated on 18/May/20 $$\mathrm{if}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right) \\ $$$$\mathrm{find}\:\mathrm{x}\: \\ $$ Answered by john santu last updated on 18/May/20…
Question Number 159845 by abdullah_ff last updated on 21/Nov/21 $$\mathrm{if}\:{q}\:=\:\mathrm{1}−{sin}\theta;\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\left({sec}\theta\:−\:{tan}\theta\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{q}} \\ $$ Commented by tounghoungko last updated on 21/Nov/21 $$\mathrm{sin}\:\theta\:=\:\mathrm{1}−{q}\: \\ $$$$\:\mathrm{cos}\:\theta\:=\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}{q}+{q}^{\mathrm{2}}…
Question Number 159787 by abdullah_ff last updated on 21/Nov/21 $$\mathrm{if}\:{cos}^{\mathrm{4}} \theta\:−\:{sin}^{\mathrm{4}} \theta\:=\:\mathrm{2}\:−\:\mathrm{5}{cos}\theta \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\theta \\ $$ Commented by cortano last updated on 21/Nov/21 $$\Rightarrow\:\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 28708 by tawa tawa last updated on 29/Jan/18 $$\mathrm{If}\:\:\:\theta\:\:=\:\:\mathrm{log}_{\mathrm{e}} \left\{\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right\}\:,\:\:\:\mathrm{prove}\:\mathrm{that}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\mathrm{tanh}\left(\mathrm{2}\theta\right)\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94241 by Ar Brandon last updated on 17/May/20 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\mathrm{arctan}\left(\mathrm{x}\right)+\mathrm{2arctan}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right)=\frac{\pi}{\mathrm{2}} \\ $$ Commented by mathmax by abdo last updated on 17/May/20…
Question Number 28648 by tawa tawa last updated on 28/Jan/18 $$\mathrm{If}\:\:\theta\:\:=\:\:\mathrm{log}_{\mathrm{e}} \left[\mathrm{tanh}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right]\:,\:\:\:\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\mathrm{3tanh}\left(\mathrm{2}\theta\right)\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28624 by Joel578 last updated on 28/Jan/18 Commented by Joel578 last updated on 28/Jan/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{given}\:\mathrm{expression} \\ $$ Commented by Tinkutara last updated on…