Question Number 194881 by tri26112004 last updated on 18/Jul/23 $${Give}\:\bigtriangleup{ABC}\: \\ $$$${Proof}:\:{sin}\:{A}\:+\:{sin}\:{B}\:+\:{sin}\:{C}\:>\:\mathrm{2} \\ $$ Answered by Frix last updated on 19/Jul/23 $$\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}. \\ $$$$\mathrm{0}<\mathrm{sin}\:{A}\:+\mathrm{sin}\:{B}\:+\mathrm{sin}\:{C}\:\:\leqslant\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 194826 by cortano12 last updated on 16/Jul/23 $$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{19}°\:=\:{p}\: \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{7}°\:=? \\ $$ Answered by dimentri last updated on 16/Jul/23 $$\:\:\:\mathrm{tan}\:\mathrm{38}°\:=\:\mathrm{tan}\:\left(\mathrm{45}°−\mathrm{7}\right) \\ $$$$\:\:\:\frac{\mathrm{2tan}\:\mathrm{19}°}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{19}°}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{7}°}{\mathrm{1}+\mathrm{tan}\:\mathrm{7}°}…
Question Number 194767 by horsebrand11 last updated on 15/Jul/23 $$\:\:\:\:\:\mathrm{tan}\:\theta\:=\:\mathrm{2}\: \\ $$$$\:\:\:\frac{\mathrm{8sin}\:\theta+\mathrm{5cos}\:\theta}{\mathrm{sin}\:^{\mathrm{3}} \theta+\mathrm{cos}\:^{\mathrm{3}} \theta+\mathrm{cos}\:\theta}\:=? \\ $$ Answered by dimentri last updated on 15/Jul/23 $$\:\:\:\mathrm{sec}\:^{\mathrm{2}} {x}=\mathrm{5}\:,\:\mathrm{tan}\:^{\mathrm{2}}…
Question Number 194766 by dimentri last updated on 15/Jul/23 $$\:\:\:\mathrm{1}+\mathrm{2cot}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\:\mathrm{3}\: \\ $$$$\:\:\:{x}=? \\ $$ Answered by Frix last updated on 15/Jul/23 $$\mathrm{1}+\mathrm{2cor}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\mathrm{3} \\ $$$$\mathrm{cot}^{\mathrm{2}} \:{x}\:=\mathrm{3}…
Question Number 194697 by cortano12 last updated on 13/Jul/23 $$\:\:\:\:\underbrace{\:} \\ $$ Answered by MM42 last updated on 13/Jul/23 $$\frac{{tanx}−{tan}\mathrm{3}{x}}{{tanx}}=\mathrm{3}\Rightarrow\frac{{tan}\mathrm{3}{x}}{{tanx}}=−\mathrm{2} \\ $$$$\frac{{cotx}}{{cotx}+{cot}\mathrm{3}{x}}=\frac{{tan}\mathrm{3}{x}}{{tanx}+{tan}\mathrm{3}{x}} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{1}−\mathrm{2}}=\mathrm{2}\:\checkmark \\…
Question Number 194426 by Mingma last updated on 06/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194327 by pete last updated on 03/Jul/23 $$\mathrm{If}\:\mathrm{ta4}\theta\:=\:\mathrm{1},\:\mathrm{find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\theta. \\ $$ Answered by MikeH last updated on 03/Jul/23 $$\mathrm{tan}\:\mathrm{4}\theta\:=\:\mathrm{1}\: \\ $$$$\Rightarrow\:\mathrm{4}\theta\:=\:\pi{n}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\pi \\ $$$$\Rightarrow\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{4}}\pi{n}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\pi \\…
Question Number 194315 by cortano12 last updated on 03/Jul/23 $$\:\:\:\:\: \\ $$ Answered by Frix last updated on 03/Jul/23 $$\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{transformed}\:\mathrm{to}\:\left[{s}=\mathrm{sin}\:{x}\right]: \\ $$$${s}^{\mathrm{4}} +\frac{{s}^{\mathrm{3}} }{\mathrm{2}}−{s}^{\mathrm{2}} −\frac{{s}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0}…
Question Number 194278 by cortano12 last updated on 02/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194304 by 281981 last updated on 02/Jul/23 Answered by cortano12 last updated on 03/Jul/23 $$\:\:\:\:\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)\:\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{sin}\:\left(\mathrm{2A}+\mathrm{2B}\right)\:\mathrm{sin}\:\left(\mathrm{2A}−\mathrm{2B}\right)=\:\mathrm{2} \\ $$$$\:\:\:\underbrace{\boldsymbol{{A}}} \\ $$ Commented by…