Question Number 87424 by Rio Michael last updated on 04/Apr/20 $$\mathrm{prove}\:\mathrm{that}\:\:\:\frac{\mathrm{sin}\:{x}\:−\:\mathrm{2}\:\mathrm{sin}\:{x}\:+\:\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{sin}\:{x}\:+\:\mathrm{2sin}\:{x}\:+\:\mathrm{sin}\:\mathrm{3}{x}}\:\equiv\:−\mathrm{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$ Commented by john santu last updated on 04/Apr/20 $$\frac{\mathrm{sin}\:\mathrm{3x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{3x}+\mathrm{3sin}\:\mathrm{x}}\:=\:\frac{\mathrm{2cos}\:\mathrm{2x}\:\mathrm{sin}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{3sin}\:\mathrm{x}} \\…
Question Number 21889 by ajfour last updated on 06/Oct/17 $${If}\:{a},{b},\:{and}\:{A}\:{of}\:{a}\:{triangle}\:\:{are}\: \\ $$$${fixed}\:{and}\:{two}\:{possible}\:{values}\:{of}\:{the}\: \\ $$$${third}\:{side}\:{be}\:{c}_{\mathrm{1}} {and}\:{c}_{\mathrm{2}} {such}\:{that} \\ $$$$\boldsymbol{{c}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{1}} \boldsymbol{{c}}_{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{2}} ^{\mathrm{2}} =\boldsymbol{{a}}^{\mathrm{2}} ,\:{then}\:{find}\:{angle}\:{A}.…
Question Number 87418 by liki last updated on 04/Apr/20 Answered by Rio Michael last updated on 04/Apr/20 $$\left(\mathrm{a}\right)\:\mathrm{since}\:\mathrm{we}\:\mathrm{have}\:\mathrm{3}\:\mathrm{match}\:\mathrm{pairs}\:\mathrm{in}\:\mathrm{the}\:\mathrm{box},\:\mathrm{then} \\ $$$$\mathrm{P}\left(\mathrm{match}\:\mathrm{pair}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}. \\ $$$$\:\left(\mathrm{b}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{six}\:\mathrm{socks}\:\mathrm{in}\:\mathrm{that}\:\mathrm{box},\:\mathrm{3}\:\mathrm{left}\:\mathrm{feet}\:\mathrm{and}\:\mathrm{3}\:\mathrm{right}\:\mathrm{fit} \\ $$$$\:\:\:\:\:\:\:\mathrm{P}\left(\mathrm{left}\:\mathrm{feet}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{and}\:\mathrm{P}\left(\mathrm{right}\:\mathrm{feet}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\…
Question Number 87419 by john santu last updated on 04/Apr/20 $$\mathrm{if}\:\mathrm{equation}\:\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{sec}\:\mathrm{x}\:−\mathrm{2tan}\:\mathrm{x}\:−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{roots}\:\mathrm{x}_{\mathrm{1}} \:\&\:\mathrm{x}_{\mathrm{2}} \:,\:\mathrm{then}\:\mathrm{the}\: \\ $$$$\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:\mathrm{x}_{\mathrm{1}} −\mathrm{cos}\:\mathrm{x}_{\mathrm{2}} \:? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{4}/\mathrm{5}\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{3}/\mathrm{4}\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{4}/\mathrm{3}\: \\ $$$$\left(\mathrm{d}\right)\:\mathrm{3}/\mathrm{2}\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{2} \\ $$…
Question Number 152949 by Riyoziyot last updated on 03/Sep/21 Answered by mr W last updated on 03/Sep/21 $${say}\:{AB}={AC}=\mathrm{1} \\ $$$${AD}={CD}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{42}} \\ $$$${BC}=\mathrm{2}\:\mathrm{cos}\:\mathrm{66} \\ $$$$\angle{BCD}=\mathrm{66}−\mathrm{42}=\mathrm{24}° \\…
Question Number 21877 by FilupS last updated on 06/Oct/17 $${x}=^{\mathrm{3}} \sqrt{\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}}+^{\mathrm{3}} \sqrt{\mathrm{7}−\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$$\: \\ $$$$\mathrm{1}.\:\mathrm{According}\:\mathrm{to}\:\mathrm{a}\:\mathrm{video},\:{x}=\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{2}.\:\mathrm{According}\:\mathrm{to}\:\mathrm{WolframAlpha}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}\approx\mathrm{0}.\mathrm{2071}+\mathrm{0}.\mathrm{3587}{i}\:\:\mathrm{for}\:“\mathrm{principal}\:\mathrm{root}'' \\ $$$$\mathrm{and}\:\:\:{x}=\mathrm{2}\sqrt{\mathrm{2}}\:\:\mathrm{for}\:“\mathrm{real}-\mathrm{valued}\:\mathrm{root}'' \\…
Question Number 152950 by john_santu last updated on 03/Sep/21 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{2}}{\mathrm{8}}+…+\frac{\mathrm{7}}{\mathrm{8}}\right)=? \\ $$ Commented by MJS_new last updated on 03/Sep/21 $$\mathrm{14} \\ $$ Terms of Service…
Question Number 87413 by ajfour last updated on 04/Apr/20 Commented by ajfour last updated on 04/Apr/20 $${Express}\:{r}_{{min}} \:{in}\:{terms}\:{of}\:{a},\:{b}\:. \\ $$ Answered by mr W last…
Question Number 21874 by tawa tawa last updated on 05/Oct/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{if}\:\:\:\mathrm{2}^{\mathrm{2006}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{17} \\ $$ Answered by mrW1 last updated on 06/Oct/17 $$\mathrm{2}^{\mathrm{2006}} =\mathrm{2}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{2004}} =\mathrm{4}×\left(\mathrm{2}^{\mathrm{4}}…
Question Number 152947 by mnjuly1970 last updated on 03/Sep/21 $$ \\ $$$$\:\:\:{prove}\::: \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \:\frac{\:{e}^{\:−\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }} }{{x}^{\:\mathrm{4}} }\:{dx}\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\Gamma\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\right) \\ $$$$ \\ $$ Answered…