Question Number 197719 by a.lgnaoui last updated on 27/Sep/23 $$\mathrm{Determiner}\:\:\:\:\boldsymbol{\mathrm{x}} \\ $$ Commented by a.lgnaoui last updated on 27/Sep/23 Commented by som(math1967) last updated on…
Question Number 197583 by cortano12 last updated on 23/Sep/23 $$\:\:\begin{cases}{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)}\:=\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{\mathrm{cos}\:\mathrm{y}}{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases} \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\: \\ $$ Answered by Sutrisno last updated on 23/Sep/23 $$\frac{\frac{{sinx}}{{cos}\left({x}+{y}\right)}}{\frac{{cosy}}{{cos}\left({x}+{y}\right)}}=\frac{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\rightarrow\frac{{sinx}}{{cosy}}=−\mathrm{1}\rightarrow{sinx}={sin}\left(\mathrm{270}^{{o}} +{y}\right)\rightarrow{x}=\mathrm{270}^{{o}} +{y} \\…
Question Number 197502 by 281981 last updated on 19/Sep/23 Commented by mokys last updated on 21/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197437 by dimentri last updated on 17/Sep/23 $$\:\:\:\:\begin{cases}{\mathrm{2sin}\:\left(\mathrm{2}{x}+{y}\right)\:\mathrm{sin}\:{y}\:=\:\mathrm{cos}\:\mathrm{2}{x}}\\{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{y}=\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\:\:{Find}\:{the}\:{solution} \\ $$ Answered by Frix last updated on 17/Sep/23 $${x}=\mathrm{tan}^{−\mathrm{1}} \:{u}\:\wedge{y}=\mathrm{tan}^{−\mathrm{1}} \:{v} \\…
Question Number 197346 by Amidip last updated on 14/Sep/23 Answered by som(math1967) last updated on 14/Sep/23 $$\:\frac{{a}+{b}}{{a}−{b}}=\frac{{tan}\left(\theta+\phi\right)}{{tan}\left(\theta−\phi\right)} \\ $$$$\:\frac{{a}+{b}}{{a}−{b}}=\frac{{sin}\left(\theta+\phi\right){cos}\left(\theta−\phi\right)}{{sin}\left(\theta−\phi\right){cos}\left(\theta+\phi\right)} \\ $$$$\frac{\mathrm{2}{a}}{\mathrm{2}{b}}=\frac{{sin}\mathrm{2}\theta}{{sin}\mathrm{2}\phi}\:\:\left[{using}\:{componendo\÷ndo}\right] \\ $$$${asin}\mathrm{2}\phi={bsin}\mathrm{2}\theta \\ $$$${a}^{\mathrm{2}}…
Question Number 197327 by mnjuly1970 last updated on 13/Sep/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{trigonometry}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{P}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{44}} {\prod}}\left(\:\:\mathrm{1}\:+\:{tan}\left({k}\right)\:\right)\:=\:?\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$ Answered…
Question Number 197073 by Abdullahrussell last updated on 07/Sep/23 Answered by Frix last updated on 07/Sep/23 $$\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:={x} \\ $$$$\mathrm{Use}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\mathrm{to}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)={x}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Now}\:\mathrm{comes}\:\mathrm{the}\:“\mathrm{trick}'' \\ $$$$\left(\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{cos}\:\frac{\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)^{\mathrm{3}}…
Question Number 196929 by Amidip last updated on 03/Sep/23 Answered by som(math1967) last updated on 03/Sep/23 $$\mathrm{2}{U}_{\mathrm{6}} −\mathrm{3}{U}_{\mathrm{4}} +\mathrm{1} \\ $$$$=\mathrm{2}\left({sin}^{\mathrm{6}} \alpha+{cos}^{\mathrm{6}} \alpha\right)−\mathrm{3}\left({sin}^{\mathrm{4}} \alpha+{cos}^{\mathrm{4}} \right)+\mathrm{1}…
Question Number 196886 by Amidip last updated on 02/Sep/23 Answered by MM42 last updated on 02/Sep/23 $${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{{p}}{{q}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{p}}{{q}−\mathrm{1}}=\frac{{sin}^{\mathrm{2}} \left(\alpha+\beta\right)}{{sin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)} \\ $$$$\left.\Rightarrow{psin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)=\left({q}−\mathrm{1}\right){sin}^{\mathrm{2}} \left(\alpha+\beta\right)\right) \\ $$$$\Rightarrow{sin}^{\mathrm{2}}…
Question Number 196881 by HeferH last updated on 02/Sep/23 Commented by MathematicalUser2357 last updated on 10/Sep/23 $$\ast\mathrm{Tears}\:\mathrm{paper}\ast \\ $$$$\mathrm{Now}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{question} \\ $$ Answered by HeferH last…