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) \\ $$$${f}_{\mathrm{4}} \left({x}\right)−{f}_{\mathrm{6}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\:^{\mathrm{4}} {x}\:+\mathrm{cos}\:^{\mathrm{4}} {x}\right)−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{cos}\:^{\mathrm{6}} {x}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{4}}…
Question Number 121705 by bemath last updated on 11/Nov/20 Answered by MJS_new last updated on 11/Nov/20 $$\mathrm{cis}\:{x}\:=\mathrm{cos}\:{x}\:+\mathrm{i}\:\mathrm{sin}\:{x}\:=\mathrm{e}^{\mathrm{i}{x}} \\ $$$$\mathrm{16}°=\frac{\mathrm{4}\pi}{\mathrm{45}};\:\mathrm{44}°=\frac{\mathrm{11}\pi}{\mathrm{45}};\:\mathrm{62}°=\frac{\mathrm{31}\pi}{\mathrm{90}} \\ $$$$\frac{\mathrm{12e}^{\frac{\mathrm{4}\pi}{\mathrm{45}}\mathrm{i}} }{\mathrm{3e}^{\frac{\mathrm{11}\pi}{\mathrm{45}}\mathrm{i}} \mathrm{2e}^{\frac{\mathrm{31}\pi}{\mathrm{90}}\mathrm{i}} }=\mathrm{2e}^{\mathrm{i}\pi\left(\frac{\mathrm{4}}{\mathrm{45}}−\frac{\mathrm{11}}{\mathrm{45}}−\frac{\mathrm{31}}{\mathrm{90}}\right)} =\mathrm{2e}^{−\mathrm{i}\frac{\pi}{\mathrm{2}}}…
Question Number 121689 by bemath last updated on 11/Nov/20 $$\:{Let}\:{a}\:{and}\:{b}\:{be}\:{real}\:{numbers}\:{such} \\ $$$${that}\:\begin{cases}{\mathrm{sin}\:{a}\:+\:\mathrm{sin}\:{b}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{cos}\:{a}\:+\:\mathrm{cos}\:{b}\:=\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}}\end{cases} \\ $$$${Evaluate}\:\mathrm{sin}\:\left({a}+{b}\right).\: \\ $$ Answered by liberty last updated on 11/Nov/20 $$\mathrm{squaring}\:\mathrm{the}\:\mathrm{both}\:\mathrm{equation}\:\mathrm{gives} \\…
Question Number 187144 by a.lgnaoui last updated on 14/Feb/23 $${determiner}\:{les}\:{elements} \\ $$$${demandes}\:\left({fig}\:{ci}−{joint}\right) \\ $$ Commented by a.lgnaoui last updated on 14/Feb/23 Terms of Service Privacy…
Question Number 121598 by TITA last updated on 09/Nov/20 $${find}\:{the}\:{solution}\:{of}\:{f}\left({x},{y},{z}\right)=\left(\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{y}\right)\left(\mathrm{tan}\:{z}\right) \\ $$$${for}\:\mathrm{0}\leqslant{x},{y},{z}\leqslant\frac{\Pi}{\mathrm{4}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121522 by benjo_mathlover last updated on 09/Nov/20 $$\mathrm{19}\:\mathrm{sin}\:\mathrm{2x}\:=\mathrm{37}\:\mathrm{cos}\:\mathrm{2x}\:+\mathrm{38}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\mathrm{x}\:. \\ $$ Commented by liberty last updated on 09/Nov/20 $$\:\frac{\mathrm{38sin}\:\mathrm{xcos}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=\:\frac{\mathrm{37}\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{cos}\:^{\mathrm{2}}…