Question Number 17921 by Tinkutara last updated on 12/Jul/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{1}°\right)\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{2}°\right)\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{3}°\right)….\left(\mathrm{3}\:−\:\mathrm{tan}^{\mathrm{2}} \mathrm{89}°\right) \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Commented by Tinkutara last updated…
Question Number 83430 by jagoll last updated on 02/Mar/20 $$\mathrm{prove}\:\mathrm{that}? \\ $$$$\mathrm{sin}\:\mathrm{3}\theta\:\mathrm{sin}^{\mathrm{3}} \:\theta\:+\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:^{\mathrm{3}} \theta\:=\:\: \\ $$$$\mathrm{cos}\:^{\mathrm{3}} \:\left(\mathrm{2}\theta\right) \\ $$ Answered by mind is power last…
Question Number 83425 by jagoll last updated on 02/Mar/20 $$\frac{\mathrm{tan}\:\mathrm{3a}}{\mathrm{tan}\:\mathrm{a}}\:=\:\mathrm{k} \\ $$$$\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{sin}\:\mathrm{3a}}{\mathrm{sin}\:\mathrm{a}}\:=\:\frac{\mathrm{2k}}{\mathrm{k}−\mathrm{1}} \\ $$ Commented by jagoll last updated on 02/Mar/20 $$\mathrm{tan}\:\mathrm{3a}\:=\:\mathrm{k}\:\mathrm{tan}\:\mathrm{a} \\ $$$$\frac{\mathrm{3tan}\:\mathrm{a}−\mathrm{tan}\:^{\mathrm{3}} \mathrm{a}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}}…
Question Number 17866 by Tinkutara last updated on 11/Jul/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sin}\:\mathrm{5}{A}\:=\:\mathrm{5}\:\mathrm{cos}^{\mathrm{4}} \:{A}\:\mathrm{sin}\:{A}\:− \\ $$$$\mathrm{10}\:\mathrm{cos}^{\mathrm{2}} \:{A}\:\mathrm{sin}^{\mathrm{3}} \:{A}\:+\:\mathrm{sin}^{\mathrm{5}} \:{A} \\ $$ Answered by alex041103 last updated on 13/Jul/17…
Question Number 83352 by jagoll last updated on 01/Mar/20 $$\sqrt{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{cos}\:\left(\mathrm{x}\right)\right)}\:=\:−\mathrm{sin}\:\left(−\mathrm{x}\right)−\mathrm{5cos}\:\left(\mathrm{x}\right) \\ $$ Commented by jagoll last updated on 01/Mar/20 $$\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}\:=\:\mathrm{sin}\:\left(\mathrm{x}\right)−\mathrm{5cos}\:\left(\mathrm{x}\right) \\ $$$$\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \:}=\:\mathrm{sin}\:\left(\mathrm{x}\right)−\mathrm{5cos}\:\left(\mathrm{x}\right) \\…
Question Number 17815 by Tinkutara last updated on 11/Jul/17 $$\mathrm{For}\:{x}\:\in\:\left(\mathrm{0},\:\pi\right),\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{sin}\:{x}\:+\:\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}\:−\:\mathrm{sin}\:\mathrm{3}{x}\:=\:\mathrm{3}\:\mathrm{has} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Infinitely}\:\mathrm{many}\:\mathrm{solutions} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Three}\:\mathrm{solutions} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{One}\:\mathrm{solution} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{No}\:\mathrm{solution} \\ $$ Commented by alex041103…
Question Number 83319 by jagoll last updated on 29/Feb/20 $$\mathrm{if}\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{tan}\:\alpha\:+\:\mathrm{tan}\:\gamma}{\mathrm{1}+\mathrm{tan}\:\alpha\mathrm{tan}\:\gamma} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{sin}\:\mathrm{2}\beta\:=\:\frac{\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\alpha\mathrm{sin}\:\mathrm{2}\gamma} \\ $$ Answered by mind is power last updated on 01/Mar/20 $${sin}\left(\mathrm{2}\beta\right)=\frac{\mathrm{2}{tg}\left(\beta\right)}{\mathrm{1}+{tg}^{\mathrm{2}} \left(\beta\right)}…
Question Number 148773 by mim24 last updated on 31/Jul/21 Answered by som(math1967) last updated on 31/Jul/21 $$\frac{\mathrm{1}}{\boldsymbol{{sin}}\mathrm{20}}\:+\frac{\sqrt{\mathrm{3}}}{\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\boldsymbol{{cos}}\mathrm{20}+\sqrt{\mathrm{3}}\boldsymbol{{sin}}\mathrm{20}}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{cos}}\mathrm{20}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{{sin}}\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}.\mathrm{2}\left(\boldsymbol{{cos}}\mathrm{60}\boldsymbol{{cos}}\mathrm{20}+\boldsymbol{{sin}}\mathrm{60}\boldsymbol{{sin}}\mathrm{20}\right)}{\mathrm{2}\boldsymbol{{sin}}\mathrm{20}\boldsymbol{{cos}}\mathrm{20}} \\ $$$$=\frac{\mathrm{4}\boldsymbol{{cos}}\left(\mathrm{60}−\mathrm{20}\right)}{\boldsymbol{{sin}}\mathrm{40}}…
Question Number 148768 by mim24 last updated on 31/Jul/21 Answered by liberty last updated on 31/Jul/21 $$\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{120}°+\mathrm{2A}\right)+\mathrm{1}−\mathrm{cos}\:\mathrm{2A}+\mathrm{1}−\mathrm{cos}\:\left(\mathrm{120}°−\mathrm{2A}\right)}{\mathrm{2}}\:= \\ $$$$\frac{\mathrm{3}−\mathrm{cos}\:\mathrm{2A}−\left\{\mathrm{cos}\:\left(\mathrm{120}°+\mathrm{2A}\right)+\mathrm{cos}\:\left(\mathrm{120}°−\mathrm{2A}\right)\right\}}{\mathrm{2}}= \\ $$$$\frac{\mathrm{3}−\mathrm{cos}\:\mathrm{2A}−\left\{\mathrm{2cos}\:\mathrm{120}°\mathrm{cos}\:\mathrm{2A}\right\}}{\mathrm{2}}= \\ $$$$\frac{\mathrm{3}−\mathrm{cos}\:\mathrm{2A}−\left\{−\mathrm{cos}\:\mathrm{2A}\right\}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$…
Question Number 148767 by mim24 last updated on 31/Jul/21 Answered by liberty last updated on 31/Jul/21 $$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\mathrm{20}°}=\frac{\mathrm{cos}\:\mathrm{20}°+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{20}°}{\mathrm{sin}\:\mathrm{20}°\mathrm{cos}\:\mathrm{20}°} \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{20}°+\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{20}°}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{40}°} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{30}°\mathrm{cos}\:\mathrm{20}°+\mathrm{cos}\:\mathrm{30}°\mathrm{sin}\:\mathrm{20}°}{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{40}°} \\ $$$$=\frac{\mathrm{4sin}\:\mathrm{50}°}{\mathrm{cos}\:\mathrm{50}°}=\mathrm{4tan}\:\mathrm{50}°\:=\:\mathrm{4cot}\:\mathrm{40}° \\ $$…