Question Number 21266 by oyshi last updated on 18/Sep/17 $$\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{tan}\:{A}}+\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{tan}\:{B}}+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{tan}\:{C}}=\mathrm{0} \\ $$ Answered by myintkhaing last updated on 18/Sep/17…
Question Number 21260 by oyshi last updated on 17/Sep/17 $${if}\:\left(\theta−\varphi\right)\:{subtle}\:{and}\:\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi=\sqrt{\mathrm{3}\left(\mathrm{cos}\:\varphi\right.} \\ $$$$\left.−\mathrm{cos}\:\theta\right) \\ $$$${so}\:{proof}\:\mathrm{sin}\:\mathrm{3}\theta+\mathrm{sin}\:\mathrm{3}\varphi=\mathrm{0} \\ $$ Answered by 951172235v last updated on 07/Feb/19 $$\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi\:=\sqrt{\mathrm{3}}\:\left(\mathrm{cos}\:\varphi−\mathrm{cos}\:\theta\right) \\…
Question Number 21259 by oyshi last updated on 17/Sep/17 $$\mathrm{2cos}\:\frac{\pi}{\mathrm{13}}\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}}=\mathrm{0} \\ $$ Answered by myintkhaing last updated on 18/Sep/17 $$\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{cos}\:\frac{\mathrm{10}\pi}{\mathrm{13}}\:+\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{13}}\:+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}\:+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}} \\ $$$$=\mathrm{cos}\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{13}}\right)+\mathrm{cos}\left(\pi−\frac{\mathrm{5}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}} \\ $$$$=\:−\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{13}}−\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{13}}+\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{13}}\:=\:\mathrm{0} \\…
Question Number 21241 by tawa tawa last updated on 17/Sep/17 $$\mathrm{Find}\:\mathrm{y}\:\mathrm{in}\:\mathrm{3rd}\:\mathrm{quadrant} \\ $$$$\mathrm{tan}\left(\mathrm{y}\:−\:\mathrm{30}\right)\:=\:\mathrm{cot}\left(\mathrm{y}\right) \\ $$ Answered by sma3l2996 last updated on 17/Sep/17 $${tan}\left({y}−\mathrm{30}\right)={cot}\left({y}\right) \\ $$$${tan}\left({y}−\mathrm{30}\right)=\frac{\mathrm{1}}{{tan}\left({y}\right)}\Leftrightarrow\frac{{tan}\left({y}\right)−{tan}\left(\mathrm{30}\right)}{\mathrm{1}+{tan}\left({y}\right){tan}\left(\mathrm{30}\right)}=\frac{\mathrm{1}}{{tany}}…
Question Number 86726 by A8;15: last updated on 30/Mar/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 86716 by john santu last updated on 30/Mar/20 $$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{17}}\right)×\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{17}}\right)\:= \\ $$ Commented by jagoll last updated on 30/Mar/20 $$\frac{\mathrm{cos}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{2sin}\:\mathrm{x}}\:×\:\mathrm{2sin}\:\mathrm{x}\:\: \\ $$$$\frac{\mathrm{sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{2sin}\:\mathrm{x}}\:=\: \\ $$$$\frac{\mathrm{sin}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{cos}\:\mathrm{8x}}{\mathrm{4sin}\:\mathrm{x}}\:=\:…
Question Number 86701 by john santu last updated on 30/Mar/20 $$\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} \\…