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Author: Tinku Tara

Question-199771

Question Number 199771 by Rupesh123 last updated on 09/Nov/23 Answered by AST last updated on 09/Nov/23 $$\frac{{sin}\left(\mathrm{4}{x}\right)=\mathrm{2}{sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{2}{x}\right)}{{a}}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{b}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}{x}\right)=\frac{{a}}{\mathrm{2}{b}}=\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} {x}=\frac{{a}+\mathrm{2}{b}}{\mathrm{4}{b}} \\ $$$$\frac{{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)}{{b}}=\frac{{sin}\left({x}\right)}{{c}}\Rightarrow{cos}\left({x}\right)=\frac{{b}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\frac{{a}+\mathrm{2}{b}}{{b}}=\frac{{b}^{\mathrm{2}}…

ABFE-Care-determiner-x-en-fonction-de-a-etb-BC-a-DE-b-

Question Number 199821 by a.lgnaoui last updated on 09/Nov/23 $$\mathrm{ABFE}\:\:\mathrm{Care} \\ $$$$\mathrm{determiner}\:\boldsymbol{\mathrm{x}}\:\mathrm{en}\:\mathrm{fonction}\:\mathrm{de}\:\mathrm{a}\:\mathrm{etb} \\ $$$$\mathrm{BC}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\mathrm{DE}=\:\boldsymbol{\mathrm{b}} \\ $$ Commented by a.lgnaoui last updated on 09/Nov/23 Commented by…

Question-199817

Question Number 199817 by ajfour last updated on 09/Nov/23 Answered by ajfour last updated on 09/Nov/23 $$\left\{{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta\right)−\frac{\mathrm{sin}\:\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}\right\}^{\mathrm{2}} \\ $$$$\:\:\:={a}\left\{{a}+\frac{\mathrm{sin}\:\theta}{\mathrm{2cos}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}\right\} \\ $$$${b}=\frac{\mathrm{sin}\:\theta}{\mathrm{2cos}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right)} \\ $$…

Question-199781

Question Number 199781 by cortano12 last updated on 09/Nov/23 $$\:\:\: \\ $$ Answered by qaz last updated on 09/Nov/23 $${f}\left({x}\right)=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\left(\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}−{x}}\right)}{\frac{\mathrm{1}}{\mathrm{1}−{x}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{{x}}+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\…

S-F-dS-F-yze-1-xze-2-xye-3-S-x-y-z-2-v-sin-u-sin-2piv-v-cos-u-2-v-sin-u-cos-2piv-2v-2-

Question Number 199736 by MathedUp last updated on 08/Nov/23 $$\int\int_{\:\boldsymbol{\mathcal{S}}} \:\hat {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\hat {\boldsymbol{\mathrm{S}}} \\ $$$$\hat {\boldsymbol{\mathrm{F}}}=−{yz}\hat {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −{xz}\hat {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −{xy}\boldsymbol{\mathrm{e}}_{\mathrm{3}} \: \\ $$$$\hat {\boldsymbol{\mathrm{S}}};\:\begin{pmatrix}{\hat {{x}}}\\{\hat…