Question Number 150670 by EDWIN88 last updated on 14/Aug/21 $$\:{Solve}\:{the}\:{equation}\: \\ $$$$\:\mathrm{sin}\:^{\mathrm{3}} {x}+\mathrm{sin}\:^{\mathrm{3}} \left(\frac{\mathrm{2}\pi}{\mathrm{3}}+{x}\right)+\mathrm{sin}\:^{\mathrm{3}} \left(\frac{\mathrm{4}\pi}{\mathrm{3}}+{x}\right)+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}{x}=\mathrm{0}\: \\ $$ Answered by EDWIN88 last updated on 14/Aug/21 Answered…
Question Number 85130 by john santu last updated on 19/Mar/20 $$\mathrm{given}\:\mathrm{f}\left(\mathrm{x}\right)=\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\mathrm{sin}\:\mathrm{x}\:+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{cos}\:\mathrm{x} \\ $$$$\mathrm{find}\:\mathrm{masimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{function} \\ $$$$\left[\mathrm{f}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \\ $$ Answered by Rio Michael last updated on 19/Mar/20…
Question Number 150609 by ajfour last updated on 13/Aug/21 $${I}\:{thought}\:{this}\:{as}\:{more}\:{basic}: \\ $$$$\frac{{sinA}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\frac{{cosA}}{{a}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{abc}} \\ $$$$\Rightarrow\:\:\boldsymbol{{tanA}}=\frac{\boldsymbol{{abc}}}{\boldsymbol{{R}}\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)} \\ $$ Terms…
Question Number 84988 by jagoll last updated on 18/Mar/20 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}\:=\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{z}+\mathrm{cot}\:^{\mathrm{2}} \mathrm{z} \\ $$ Commented by MJS last updated on 18/Mar/20…
Question Number 19385 by NEC last updated on 10/Aug/17 $${tan}^{\mathrm{2}} \beta=−\mathrm{1} \\ $$$$ \\ $$$${find}\:\beta….\:{lets}\:{solve}\:{for}\:{fun} \\ $$ Commented by NEC last updated on 10/Aug/17 $${please}\:{help}…
Question Number 19367 by Tinkutara last updated on 10/Aug/17 $$\mathrm{If}\:\alpha,\:\beta,\:\gamma\:\mathrm{and}\:\delta\:\mathrm{are}\:\mathrm{four}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{tan}\:\left(\theta\:+\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right)\:=\:\mathrm{3}\:\mathrm{tan}\:\mathrm{3}\theta,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\Sigma\mathrm{tan}\:\alpha\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:=\:−\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:=\:−\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:\mathrm{tan}\:\delta\:=\:−\mathrm{3} \\ $$ Answered by ajfour…
Question Number 84868 by jagoll last updated on 17/Mar/20 $$\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{x}\right)}\:=\:−\mathrm{cos}\:\mathrm{x}+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{x}−\pi\right) \\ $$ Answered by john santu last updated on 17/Mar/20 $$\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{x}\right)\:=\:−\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{sin}\:\left(\mathrm{x}−\pi\right)\:=\:−\mathrm{sin}\:\mathrm{x} \\…
Question Number 19292 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{equality} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:…\:\mathrm{sin}\:\frac{\left({n}\:−\:\mathrm{1}\right)\pi}{\mathrm{2}{n}}\:=\:\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$ Commented by prakash jain last updated on 09/Aug/17 $$\mathrm{you}\:\mathrm{can}\:\mathrm{try}\:\mathrm{with}\:\mathrm{the}\:\mathrm{following} \\…
Question Number 19291 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°}\:=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:. \\ $$ Answered by Tinkutara last updated on 16/Aug/17 $$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:+\:\frac{\mathrm{sin}\:\mathrm{48}°\:+\:\mathrm{sin}\:\mathrm{24}°}{\mathrm{sin}\:\mathrm{48}°\:\mathrm{sin}\:\mathrm{24}°} \\…
Question Number 19243 by Tinkutara last updated on 07/Aug/17 $$\mathrm{Let}\:{a},\:{b},\:{c}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{opposite}\:\mathrm{the} \\ $$$$\mathrm{angles}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{respectively}\:\mathrm{of}\:\mathrm{a} \\ $$$$\Delta\mathrm{ABC}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{a}\:+\:{b}\:=\:{kc} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\:+\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:=\:{k}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}. \\ $$ Commented by Tinkutara last updated…