Question Number 101554 by student work last updated on 03/Jul/20 $$\frac{\mathrm{1}}{\mathrm{cos80}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{sin80}}=? \\ $$ Answered by bramlex last updated on 03/Jul/20 $$\frac{\mathrm{sin}\:\mathrm{80}^{\mathrm{o}} −\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{80}^{\mathrm{o}} }{\mathrm{sin}\:\mathrm{80}^{\mathrm{o}} \mathrm{cos}\:\mathrm{80}^{\mathrm{o}} }\:=\:…
Question Number 101522 by bramlex last updated on 03/Jul/20 $$\mathrm{calculate}\::\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{47}^{\mathrm{o}} +\mathrm{cos}\:^{\mathrm{2}} \mathrm{73}^{\mathrm{o}} +\mathrm{cos}\:\mathrm{47}^{\mathrm{o}} .\mathrm{cos}\:\mathrm{73}^{\mathrm{o}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by bemath last updated on 03/Jul/20…
Question Number 35951 by gyugfeet last updated on 26/May/18 $$\left({cos}\theta+{cos}\beta/{sin}\theta−{sin}\beta\right)=\left({sin}\theta+{sin}\beta/{cos}\theta−{cos}\beta\right)\:{prove}\:{ghis} \\ $$ Commented by Rasheed.Sindhi last updated on 27/May/18 $$\mathcal{D}\mathrm{o}\:\mathrm{you}\:\mathrm{mean} \\ $$$$\mathrm{cos}\theta+\frac{\mathrm{cos}\beta}{\mathrm{sin}\theta}−\mathrm{sin}\beta=\mathrm{sin}\theta+\frac{\mathrm{sin}\beta}{\mathrm{cos}\theta}−\mathrm{cos}\beta \\ $$$$\mathrm{or} \\…
Question Number 35886 by behi83417@gmail.com last updated on 25/May/18 Answered by ajfour last updated on 25/May/18 $$\mathrm{sin}\:\left({x}+{y}\right)−\mathrm{sin}\:{x}\:=\:\mathrm{sin}\:{y} \\ $$$$\mathrm{2cos}\:\left({x}+\frac{{y}}{\mathrm{2}}\right)\mathrm{sin}\:\frac{{y}}{\mathrm{2}}=\mathrm{2cos}\:\frac{{y}}{\mathrm{2}}\mathrm{sin}\:\frac{{y}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\frac{{y}}{\mathrm{2}}\left[\mathrm{cos}\:\left({x}+\frac{{y}}{\mathrm{2}}\right)−\mathrm{cos}\:\frac{{y}}{\mathrm{2}}\right]=\mathrm{0} \\ $$$${or}\:\:\left(\mathrm{sin}\:\frac{{y}}{\mathrm{2}}\right)\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}+{y}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{y}=\mathrm{2}{m}\pi\:,\:{or}\:\:{x}=\mathrm{2}{n}\pi,\:{or}…
Question Number 101382 by student work last updated on 02/Jul/20 Answered by Rio Michael last updated on 02/Jul/20 $$\mathrm{we}\:\mathrm{know}\:\mathrm{or}\:\mathrm{we}\:\mathrm{are}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:\:\mathrm{sin}\:{A}\:+\:\mathrm{sin}{B}\:+\:\mathrm{sin}\:{C}\:=\:\mathrm{4}\:\mathrm{cos}\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\frac{{C}}{\mathrm{2}}\:\left(\left(\mathrm{1}\right)\right. \\ $$$$\mathrm{but}\:{A}\:+\:{B}\:+\:{C}\:=\:\pi\left(\mathrm{180}°\right) \\ $$$$\Rightarrow\:{A}\:+\:{B}\:=\:\pi−{C}\:\mathrm{and}\:\:\frac{{A}\:+{B}}{\mathrm{2}}\:=\:\frac{\pi\:−\:{C}}{\mathrm{2}} \\ $$$$…
Question Number 35844 by behi83417@gmail.com last updated on 24/May/18 Commented by abdo mathsup 649 cc last updated on 30/May/18 $${let}\:{put}\:\theta\:={arctant}\:\:{with}\:{t}>\mathrm{0} \\ $$$$\left({e}\right)\Leftrightarrow\:\mathrm{1}+{sin}\left({arctant}\right)\:=\mathrm{4}{t}\:\Leftrightarrow \\ $$$$\mathrm{1}+\:\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\mathrm{4}{t}\:\Leftrightarrow\:{t}\:+\sqrt{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 166774 by BagusSetyoWibowo last updated on 27/Feb/22 Commented by cortano1 last updated on 27/Feb/22 $$\Rightarrow\mathrm{x}=\sqrt{\mathrm{25}−\mathrm{24}.\frac{\mathrm{1}}{\mathrm{2}}}=\sqrt{\mathrm{13}}\:\mathrm{cm} \\ $$$$\Rightarrow\mathrm{Area}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}=\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{cm}^{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 35654 by chakraborty ankit last updated on 21/May/18 $${if}\:\:\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta=\mathrm{tan}\:^{\mathrm{2}} \emptyset\:\:{Then}\:{proof}\:{that} \\ $$$$\mathrm{2cos}\:^{\mathrm{2}} \emptyset−\mathrm{1}=\mathrm{cos}\:^{\mathrm{2}} \emptyset−\mathrm{sin}\:^{\mathrm{2}} \emptyset=\mathrm{2tan}\:^{\mathrm{2}} \theta \\ $$ Answered by tanmay.chaudhury50@gmail.com…
Question Number 166719 by mathlove last updated on 26/Feb/22 $$\mathrm{sin}\:{x}\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}={a} \\ $$$${any}\:{price}\:{to}\:{a} \\ $$$${is}\:{corect}\:{the}\:{answer}\:\:\:\:{a}=? \\ $$ Commented by cortano1 last updated on 26/Feb/22 $$\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{2x}\:=\:\mathrm{2}{a} \\…
Question Number 166718 by mathlove last updated on 26/Feb/22 $${csc}\mathrm{10}−\sqrt{\mathrm{3}}{sec}\mathrm{10}=? \\ $$ Commented by cortano1 last updated on 26/Feb/22 $$\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\mathrm{10}°}\:=\:\frac{\mathrm{cos}\:\mathrm{10}°−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{10}°}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\:\frac{\mathrm{2}\left(\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{10}°−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\frac{\mathrm{4}\left(\mathrm{cos}\:\mathrm{60}°\:\mathrm{cos}\:\mathrm{10}°−\mathrm{sin}\:\mathrm{60}°\:\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\…