Question Number 189270 by Rupesh123 last updated on 14/Mar/23 Commented by maths_plus last updated on 14/Mar/23 $$\mathrm{good} \\ $$ Answered by HeferH last updated on…
Question Number 189256 by BaliramKumar last updated on 13/Mar/23 $${Prove}\:{that} \\ $$$$\mathrm{sin10}°\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−………..\infty}}}}}}}}}}}}} \\ $$ Answered by Frix last updated on 14/Mar/23 $$\mathrm{sin}\:\mathrm{10}°\:=\mathrm{sin}\:\frac{\pi}{\mathrm{18}} \\ $$$$\mathrm{First}\:\mathrm{let}'\mathrm{s}\:\mathrm{find}\:{x}=\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{x}}}} \\…
Question Number 189233 by a.lgnaoui last updated on 13/Mar/23 $$\mathrm{1}\bullet{Evaluer}\::\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right) \\ $$$$\mathrm{2}\bullet{En}\:{deduire}:\frac{\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right)}{\boldsymbol{{Aire}}\left(\boldsymbol{{ABCD}}\right)} \\ $$ Commented by a.lgnaoui last updated on 13/Mar/23 $$\left[\boldsymbol{{A}}\:'\boldsymbol{{B}}'\:\boldsymbol{{D}}\right]{alignes} \\ $$ Commented…
Question Number 123641 by benjo_mathlover last updated on 27/Nov/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 189125 by Rupesh123 last updated on 12/Mar/23 Answered by cortano12 last updated on 12/Mar/23 $$\:\mathrm{cos}\:\mathrm{18}°=\mathrm{2cos}^{\mathrm{2}} \:\mathrm{9}°−\mathrm{1} \\ $$$$\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{9}°=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{18}°}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{18}°=\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{18}°} \\…
Question Number 57959 by rahul 19 last updated on 15/Apr/19 Answered by MJS last updated on 16/Apr/19 $$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:={y}\:\Rightarrow\:{x}=\mathrm{sin}\:\frac{{y}}{\mathrm{3}} \\ $$$$\mathrm{with}\:{x}\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\wedge\:{y}\in\left[−\frac{\mathrm{3}\pi}{\mathrm{2}};\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:\mathrm{increases}\:\mathrm{for}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\…
Question Number 189026 by Rupesh123 last updated on 10/Mar/23 Answered by Ar Brandon last updated on 10/Mar/23 $$\mathrm{cot}^{\mathrm{2}} {x}+\mathrm{1}=\mathrm{cosec}^{\mathrm{2}} {x}\: \\ $$$$\mid\mathrm{4}−{a}\mid+\mathrm{1}=\frac{\mid{a}\mid}{\mathrm{3}}\:,\:\mathrm{0}<{a}\leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}−{a}+\mathrm{1}=\frac{{a}}{\mathrm{3}}\:\Rightarrow{a}=\frac{\mathrm{15}}{\mathrm{4}} \\…
Question Number 189012 by TUN last updated on 10/Mar/23 $${Triangle}\:{ABC}\:{have}:\: \\ $$$${sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}=\sqrt{\mathrm{3}}\left({cosA}+{cosB}+{cosC}\right) \\ $$$$=>\:{Prove}\:{that}\:{ABC}\:{is}\:{equilateral}\:{triangle} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 57915 by Tinkutara last updated on 14/Apr/19 Answered by mr W last updated on 14/Apr/19 $$\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)+\mathrm{2}^{\mathrm{tan}^{\mathrm{2}} \:\theta} =\mathrm{0} \\ $$$${let}\:{t}=\mathrm{tan}^{\mathrm{2}} \:\theta\geqslant\mathrm{0}…
Question Number 188912 by Mingma last updated on 08/Mar/23 Answered by a.lgnaoui last updated on 09/Mar/23 $${posons}\:\:\frac{\theta}{\mathrm{3}}={x} \\ $$$$\mathrm{sin}\:\left(\frac{\pi−\theta}{\mathrm{3}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{sin}\:\left(\frac{\pi+\theta}{\mathrm{3}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right) \\ $$$$\:\: \\ $$$$\mathrm{sin}^{\mathrm{3}}…