Question Number 86701 by john santu last updated on 30/Mar/20 $$\mathrm{find}\:\mathrm{solution}\: \\ $$$$\mathrm{4}^{\mathrm{sin}\:\mathrm{x}\:−\frac{\mathrm{1}}{\mathrm{4}}} \:−\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:.\mathrm{2}^{\mathrm{sin}\:\mathrm{x}} \:−\mathrm{1}\:=\:\mathrm{0}\: \\ $$$$\mathrm{in}\:\mathrm{x}\:\in\left[\:\mathrm{0},\mathrm{2}\pi\:\right]\: \\ $$ Commented by john santu last updated…
Question Number 21141 by oyshi last updated on 14/Sep/17 $${if}\:\mathrm{sin}\:\alpha\mathrm{sin}\:\beta−\mathrm{cos}\:\alpha\mathrm{cos}\:\beta+\mathrm{1}=\mathrm{0}\: \\ $$$${so}\:{proof}\:{that}\:\mathrm{1}+\mathrm{cot}\:\alpha\mathrm{tan}\:\beta=\mathrm{0} \\ $$ Answered by $@ty@m last updated on 15/Sep/17 $$\mathrm{sin}\:\alpha\mathrm{sin}\:\beta−\mathrm{cos}\:\alpha\mathrm{cos}\:\beta+\mathrm{1}=\mathrm{0}\: \\ $$$$\mathrm{cos}\:\alpha\mathrm{cos}\:\beta−\mathrm{sin}\:\alpha\mathrm{sin}\:\beta=\mathrm{1} \\…
Question Number 21142 by oyshi last updated on 14/Sep/17 $${if}\:{A}+{B}=\frac{\pi}{\mathrm{4}} \\ $$$${so}\:{proof}\:\left(\mathrm{1}+\mathrm{tan}\:{A}\right)\left(\mathrm{1}+\mathrm{tan}\:{B}\right)=\mathrm{2} \\ $$ Commented by dioph last updated on 14/Sep/17 $$\mathrm{tan}\:{A}+{B}\:=\:\mathrm{1} \\ $$$$\frac{\mathrm{tan}\:{A}\:+\:\mathrm{tan}\:{B}}{\mathrm{1}\:−\:\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}}\:=\:\mathrm{1} \\…
Question Number 21140 by oyshi last updated on 14/Sep/17 $${if}\:\mathrm{tan}\:\beta=\frac{\mathrm{2sin}\:\alpha\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\alpha+\gamma\right)} \\ $$$${so}\:{proof}\:\mathrm{cot}\:\gamma+\mathrm{cot}\:\alpha=\mathrm{2cot}\:\beta \\ $$ Commented by $@ty@m last updated on 15/Sep/17 $${Step}\:\mathrm{1}.\:{Take}\:{reciprocal}\:{of}\:{both}\:{sides}. \\ $$$${Step}\:\mathrm{2}.\:{Multiply}\:{both}\:{sides}\:{by}\:\mathrm{2} \\…
Question Number 152175 by john_santu last updated on 26/Aug/21 $$\mathrm{what}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{if}\:\mathrm{given} \\ $$$$\:\mathrm{cos}\:\mathrm{3x}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$ Commented by MJS_new last updated on 26/Aug/21 $$\mathrm{cos}\:\mathrm{3}{x}\:=\mathrm{4cos}^{\mathrm{3}} \:{x}\:−\mathrm{3cos}\:{x} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=\pm\mathrm{2cos}\:{x}\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}}…
ABC-is-an-isocel-triangle-such-as-AB-AC-3-and-BC-4-and-are-its-angles-Show-that-cos-2-sin-2-Hi-sirs-
Question Number 86614 by mathocean1 last updated on 29/Mar/20 $${ABC}\:{is}\:{an}\:{isocel}\:{triangle}\:{such}\:{as} \\ $$$${AB}={AC}=\mathrm{3}\:\:{and}\:{BC}=\mathrm{4} \\ $$$$\alpha\:,\:\beta\:,\:{and}\:\gamma\:{are}\:{its}\:{angles}. \\ $$$${Show}\:{that}\:{cos}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)={sin}\left(\frac{\gamma}{\mathrm{2}}\right) \\ $$$${Hi}\:{sirs}… \\ $$ Answered by TANMAY PANACEA. last…
Question Number 21060 by NECx last updated on 11/Sep/17 $$\mathrm{write}\:\mathrm{sin}\:\mathrm{1}°\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{show}\:\mathrm{workings}. \\ $$ Commented by NECx last updated on 11/Sep/17…
Question Number 21050 by Tinkutara last updated on 10/Sep/17 $$\mathrm{The}\:\mathrm{most}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{sin}{x}\:+\:\mathrm{cos}{x}\:=\:\underset{{a}\in{R}} {\mathrm{min}}\left\{\mathrm{1},\:{a}^{\mathrm{2}} \:−\:\mathrm{4}{a}\:+\:\mathrm{6}\right\} \\ $$$$\mathrm{is} \\ $$ Answered by mrW1 last updated on 11/Sep/17…
Question Number 86550 by DuDono last updated on 29/Mar/20 $$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{tan}\:^{−\mathrm{1}} {x}=\frac{\mathrm{ln}\:\left(−{x}^{\mathrm{2}} +\mathrm{2}{ix}+\mathrm{1}\right)−\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}{i}} \\ $$ Commented by john santu last updated on 29/Mar/20…
Question Number 86540 by jagoll last updated on 29/Mar/20 $$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{B}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\:=\: \\ $$$$\mathrm{4}\:\mathrm{cos}\:\left(\frac{\pi+\mathrm{A}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\pi+\mathrm{B}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\pi−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{where}\:\mathrm{A}+\mathrm{B}+\mathrm{C}\:=\:\pi \\ $$ Commented by jagoll last updated on 29/Mar/20…