Question Number 197073 by Abdullahrussell last updated on 07/Sep/23 Answered by Frix last updated on 07/Sep/23 $$\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:={x} \\ $$$$\mathrm{Use}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\mathrm{to}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)={x}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Now}\:\mathrm{comes}\:\mathrm{the}\:“\mathrm{trick}'' \\ $$$$\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)^{\mathrm{3}}…
Question Number 197064 by Amidip last updated on 07/Sep/23 Answered by witcher3 last updated on 07/Sep/23 $$\mathrm{x}\sqrt{\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }+\mathrm{y}\left(\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{a} \\ $$$$\mathrm{a}^{\mathrm{2}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}}…
Question Number 197099 by Erico last updated on 07/Sep/23 $$\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{cos}{t}\right)}{\mathrm{sin}{t}}\:\mathrm{d}{t}=??? \\ $$ Answered by witcher3 last updated on 07/Sep/23 $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\left(\mathrm{t}\right)\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{t}\right)\right)}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{t}\right)}\mathrm{dt},\mathrm{cos}\left(\mathrm{t}\right)=\mathrm{y}…
Question Number 197060 by universe last updated on 07/Sep/23 $$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}+\alpha\mathrm{sin}{t}\right)}{\mathrm{sin}{t}}{dt}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arccos}\alpha\right)^{\mathrm{2}} \\ $$ Commented by universe last updated on 07/Sep/23 $${question}\:\mathrm{196950}…
Question Number 197094 by MrGHK last updated on 07/Sep/23 Answered by sniper237 last updated on 07/Sep/23 $$\overset{{schwarz}\:{theo}} {\Rightarrow}\:\partial_{{x}} \left(\:{t}\partial_{{t}} {u}+\mathrm{2}{u}\right)={x}^{\mathrm{2}} \\ $$$$\overset{\int{dx}} {\Rightarrow}{t}\partial_{{t}} {u}+\mathrm{2}{u}\:=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{f}\left({t}\right)…
Question Number 197095 by Mingma last updated on 07/Sep/23 Answered by mahdipoor last updated on 07/Sep/23 $$\frac{{AB}}{{sin}\mathrm{4}{a}}=\frac{{BM}}{{sina}}\:\:\: \\ $$$$\:\frac{{CD}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sin}\mathrm{2}{a}}\Rightarrow\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sina}}\Rightarrow \\ $$$$\frac{{MC}}{{sina}}+\frac{{BM}}{{sina}}=\frac{{CB}}{{sina}}=\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}+\frac{{AB}}{{sin}\mathrm{4}{a}} \\ $$$$\Rightarrow\frac{{sin}\mathrm{4}{a}}{{sina}}=\mathrm{4}{cos}\mathrm{2}{a}.{cosa}=\mathrm{2}{cosa}+\mathrm{1}\Rightarrow \\ $$$$\mathrm{8}{cos}^{\mathrm{3}}…
Question Number 197063 by otchereabdullai@gmail.com last updated on 07/Sep/23 Answered by MM42 last updated on 07/Sep/23 Commented by MM42 last updated on 07/Sep/23 $$\mid{A}−\left({B}\cup{C}\right)\mid=\mathrm{7}\:\:\&\:\:\mid{C}\mid=\mathrm{55}\:\checkmark \\…
Question Number 197089 by pete last updated on 07/Sep/23 $$\mathrm{Simplify}\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{i}}\right)^{\mathrm{10}} \\ $$ Answered by JDamian last updated on 07/Sep/23 $$\left(\frac{{z}}{{z}^{\ast} }\right)^{{n}} =\left(\frac{\cancel{\mid{z}\mid}\centerdot{e}^{{i}\varphi} }{\cancel{\mid{z}\mid}\centerdot{e}^{−{i}\varphi} }\right)^{{n}} ={e}^{{i}\mathrm{2}{n}\varphi}…
Question Number 197057 by ajfour last updated on 07/Sep/23 Commented by ajfour last updated on 07/Sep/23 $${The}\:{ant}\:{has}\:{to}\:{climb}\:{up}\:{the}\:{plane} \\ $$$${and}\:{surmount}\:{the}\:{wall}\:{of}\:{height}\:{c}, \\ $$$${and}\:{descend}\:{then}\:{reach}\:{B}.\:{Find}\:{the} \\ $$$${shortest}\:{length}\:{of}\:{path}. \\ $$…
Question Number 197017 by sonukgindia last updated on 06/Sep/23 Commented by mr W last updated on 26/Sep/23 $${see}\:{Q}#\mathrm{197670} \\ $$ Terms of Service Privacy Policy…