Question Number 56681 by Joel578 last updated on 21/Mar/19 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{sin}\:\mid{x}\mid\:\leqslant\:\mid{x}\mid\:\leqslant\:\mathrm{tan}\:\mid{x}\mid\:\:\:\:\mathrm{for}\:\:\:\:\mid{x}\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$ Commented by Abdo msup. last updated on 22/Mar/19 $${let}\:\mid{x}\mid={t}\:\:{let}\:{prove}\:{that}\:{sint}\leqslant{t}\leqslant{tant}\:{for}\mathrm{0}\leqslant{t}<\frac{\pi}{\mathrm{2}} \\ $$$${let}\:{W}\left({x}\right)={t}−{sint}\:\Rightarrow{W}^{,}…
Question Number 122118 by AbdullahMohammadNurusSafa last updated on 14/Nov/20 $$\frac{{tanA}}{\mathrm{1}−{cotA}}\:+\:\frac{{cotA}}{\mathrm{1}−{tanA}}\:=\:{secA}\:{cosecA}\:+\mathrm{1} \\ $$$$\boldsymbol{{Please}}\:\boldsymbol{{prove}}\:\boldsymbol{{it}}\:\boldsymbol{{for}}\:\boldsymbol{{me}}!!! \\ $$ Answered by MJS_new last updated on 14/Nov/20 $$\frac{{t}}{\mathrm{1}−\frac{\mathrm{1}}{{t}}}+\frac{\frac{\mathrm{1}}{{t}}}{\mathrm{1}−{t}}=\frac{{t}^{\mathrm{2}} }{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}\left({t}−\mathrm{1}\right)}={t}+\frac{\mathrm{1}}{{t}}+\mathrm{1}= \\ $$$$=\frac{{s}}{{c}}+\frac{{c}}{{s}}+\mathrm{1}=\frac{{s}^{\mathrm{2}}…
Question Number 122076 by peter frank last updated on 14/Nov/20 Answered by MJS_new last updated on 14/Nov/20 $$\mathrm{let}\:{t}=\mathrm{tan}\:{A} \\ $$$$\frac{\mathrm{1}+{t}−\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}+{t}−\mathrm{1}}=\frac{\mathrm{1}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}−{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}+{t}+\mathrm{1}} \\…
Question Number 122053 by 676597498 last updated on 13/Nov/20 $$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{E}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)=? \\ $$$$\mathrm{E}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{common} \\ $$$$\mathrm{interger}\:\mathrm{of}\:\mathrm{x} \\ $$ Commented by mr W last updated on 13/Nov/20 $${x}−\left[{x}\right]=\left\{{x}\right\}…
Question Number 187559 by Mingma last updated on 18/Feb/23 Answered by cortano12 last updated on 19/Feb/23 $$\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{y}}\:.\frac{\mathrm{cos}\:{y}}{\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\frac{\mathrm{tan}\:{x}}{\mathrm{tan}\:{y}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\mathrm{3tan}\:{x}=\mathrm{tan}\:{y} \\ $$$$\Rightarrow\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{4tan}\:{x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}}\:=\pm\:\sqrt{\mathrm{15}} \\ $$$$\Rightarrow\mathrm{2sin}\:{x}=\mathrm{sin}\:{y}\:;\:\mathrm{2cos}\:{x}=\mathrm{3cos}\:{y} \\…
Question Number 187548 by Rupesh123 last updated on 18/Feb/23 Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23 $${z}^{\mathrm{9}} +{z}^{\mathrm{6}} +{z}^{\mathrm{3}} =−\mathrm{1}\:\Rightarrow{z}^{\mathrm{9}} +{z}^{\mathrm{6}} +{z}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{z}^{\mathrm{12}}…
Question Number 187491 by cortano12 last updated on 18/Feb/23 $$\:\:\begin{cases}{\mathrm{sin}\:{x}+\mathrm{sin}\:{y}={a}}\\{\mathrm{cos}\:{x}+\mathrm{cos}\:{y}={b}}\end{cases} \\ $$$$\:\:\:\:\mathrm{tan}\:{x}+\mathrm{tan}\:{y}=? \\ $$ Answered by horsebrand11 last updated on 18/Feb/23 $$\:\left(\ast\right)\:\mathrm{tan}\:\left(\frac{{x}−{y}}{\mathrm{2}}\right)=\frac{{a}}{{b}} \\ $$$$\:\:\Rightarrow\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}}…
Question Number 187376 by Rupesh123 last updated on 16/Feb/23 Answered by Frix last updated on 17/Feb/23 $${xy}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}+{y}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\sqrt{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$$\mathrm{Solve}\:\mathrm{this}\:\left[\mathrm{which}\:\mathrm{is}\:\mathrm{easy}\right]\:\Rightarrow \\ $$$$\alpha=\mathrm{sin}^{−\mathrm{1}}…
Question Number 187375 by Rupesh123 last updated on 16/Feb/23 Commented by mokys last updated on 16/Feb/23 $$\frac{\frac{{cosx}−{sinx}}{{cosx}}}{\frac{{sinx}−{cosx}}{{sinx}}}\:\:=\:\frac{−{secx}}{{cscx}}\:=\:−{tanx}\:\neq\:\mathrm{2}{sinx} \\ $$ Commented by Frix last updated on…
Question Number 187311 by anurup last updated on 16/Feb/23 $$\mathrm{If}\:{f}_{{k}} \left({x}\right)=\frac{\mathrm{1}}{{k}}\left(\mathrm{sin}\:^{{k}} {x}+\mathrm{cos}\:^{{k}} {x}\right)\:\mathrm{find}\:{f}_{\mathrm{4}} \left({x}\right)−{f}_{\mathrm{6}} \left({x}\right) \\ $$$${f}_{\mathrm{4}} \left({x}\right)−{f}_{\mathrm{6}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\:^{\mathrm{4}} {x}\:+\mathrm{cos}\:^{\mathrm{4}} {x}\right)−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{cos}\:^{\mathrm{6}} {x}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{4}}…