Question Number 15641 by tawa tawa last updated on 12/Jun/17 $$\mathrm{Show}\:\mathrm{that}. \\ $$$$\sqrt{\mathrm{3}}\left(\mathrm{cosec20}\right)\:−\:\mathrm{sec20}\:=\:\mathrm{4} \\ $$ Answered by mrW1 last updated on 12/Jun/17 $$\frac{\sqrt{\mathrm{3}}}{\mathrm{sin}\:\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}}=\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{20}−\mathrm{1sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{20cos}\:\mathrm{20}} \\ $$$$=\frac{\mathrm{4}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\mathrm{20}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}\right)}{\mathrm{sin}\:\mathrm{40}}…
Question Number 81170 by jagoll last updated on 10/Feb/20 $${if}\:\mathrm{cos}\:^{\mathrm{3}} {x}+\mathrm{cos}\:^{−\mathrm{3}} {x}\:\:=\mathrm{0} \\ $$$${find}\:\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x} \\ $$ Commented by jagoll last updated on 10/Feb/20 $${i}'{m}\:{find}\:\mathrm{cos}\:{x}=\:\sqrt[{\mathrm{3}\:}]{\:\pm{i}\:} \\…
Question Number 81149 by TawaTawa last updated on 09/Feb/20 Answered by mind is power last updated on 09/Feb/20 $${tan}\left({B}\right)=\frac{{AC}}{{AB}} \\ $$$${tg}\left({B}−\theta\right)=\frac{{AD}}{{AB}}\Rightarrow\frac{{tg}\left({B}\right)}{{tg}\left({B}−\theta\right)}=\frac{{AC}}{{AD}}=\mathrm{4} \\ $$$$\Rightarrow{tg}\left({B}\right)=\mathrm{4}{tg}\left({B}−\theta\right)\Rightarrow{tg}\left({B}\right)=\mathrm{4}\frac{{tg}\left({B}\right)−{tg}\left(\theta\right)}{\mathrm{1}+{tg}\left({B}\right){tg}\left(\theta\right)} \\ $$$$\Rightarrow\mathrm{3}{tg}\left({B}\right)={tg}\left(\theta\right)\left({tg}^{\mathrm{2}}…
Question Number 15578 by Joel577 last updated on 12/Jun/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cot}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{14}}\right)\:+\:\mathrm{cot}^{\mathrm{2}} \:\left(\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\:+\:\mathrm{cot}^{\mathrm{2}} \:\left(\frac{\mathrm{5}\pi}{\mathrm{14}}\right)\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 15555 by Tinkutara last updated on 11/Jun/17 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{cos}\:\mathrm{8}{x}\:−\:\mathrm{cos}\:\mathrm{7}{x}}{\mathrm{1}\:+\:\mathrm{2}\:\mathrm{cos}\:\mathrm{5}{x}}\:=\:\mathrm{cos}\:\mathrm{3}{x}\:−\:\mathrm{cos}\:\mathrm{2}{x} \\ $$ Answered by ajfour last updated on 11/Jun/17 $${considering} \\ $$$${f}\left({x}\right)=\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{5}{x}\right)\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{2}{x}\right) \\…
Question Number 81062 by john santu last updated on 09/Feb/20 $$\mathrm{cos}\:{x}\:\sqrt{\mathrm{1}+\mathrm{sin}\:{x}−\mathrm{2cos}\:{x}}\:=\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x} \\ $$$${in}\:\left[\:−\mathrm{5}\pi\:,\:−\frac{\mathrm{7}\pi}{\mathrm{2}}\right]\: \\ $$ Commented by MJS last updated on 09/Feb/20 $$\mathrm{sorry}\:\mathrm{no}\:\mathrm{time}\:\mathrm{today} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{let}\:{x}=\mathrm{2arctan}\:{t}…
Question Number 81057 by mathocean1 last updated on 09/Feb/20 $${solve}\:\mathrm{in}\:\left[−\pi;\pi\right]\: \\ $$$$\left({E}\right):\:{sin}\mathrm{3}{x}=−{sin}\mathrm{2}{x} \\ $$ Commented by jagoll last updated on 09/Feb/20 $$\mathrm{sin}\:\mathrm{3}{x}+\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{0} \\ $$$$\mathrm{2sin}\:\left(\frac{\mathrm{5}{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{0} \\…
Question Number 15471 by Mr easymsn last updated on 10/Jun/17 Answered by sma3l2996 last updated on 10/Jun/17 $$\left({a}\right) \\ $$$$\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)+{sinx}=\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} {x}−{cosx} \\ $$$${sin}\left({x}\right)\left(\mathrm{2}{cos}\left({x}\right)+\mathrm{1}\right)=\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} {x}−{cosx} \\…
Question Number 80998 by Power last updated on 08/Feb/20 Commented by mr W last updated on 08/Feb/20 $${just}\:{as}\:{for}\:{a}\:{lot}\:{of}\:{your}\:{other}\:{questions}, \\ $$$${you}\:{also}\:{need}\:{a}\:{calculator}\:{for}\:{this}\:{one}. \\ $$ Answered by jagoll…
Question Number 81003 by mathocean1 last updated on 08/Feb/20 $$\left(\mathrm{E}\right):\:\mathrm{sin2}{x}={cosx}+{sinx}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}.\:{show}\:{that}\:\left(\mathrm{E}\right)\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\: \\ $$$$\left(\mathrm{E}'\right):\:\mathrm{2cos}^{\mathrm{2}} \mathrm{X}−\sqrt{\mathrm{2}}\mathrm{cosX}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{with}\:\mathrm{X}={x}−\frac{\pi}{\mathrm{4}}. \\ $$ Commented by MJS last updated on…