Question Number 8267 by lepan last updated on 04/Oct/16 $${Show}\:{that}\:{sinA}+{sinB}=\mathrm{2}{sin}\frac{{A}+{B}}{\mathrm{2}}\:{cos}\frac{{A}−{B}}{\mathrm{2}}. \\ $$$$ \\ $$ Answered by Yozzias last updated on 04/Oct/16 $$\mathrm{Let}\:\mathrm{p}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{and}\:\mathrm{q}=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}−\mathrm{B}\right). \\ $$$$\therefore\:\mathrm{2p}=\mathrm{A}+\mathrm{B}\:\mathrm{and}\:\mathrm{2q}=\mathrm{A}−\mathrm{B}. \\…
Question Number 8257 by lepan last updated on 04/Oct/16 $${If}\:{A}+{B}+{C}=\mathrm{90}°\:,{show}\:{that}\: \\ $$$${tanA}\:{tanB}+{tanB}\:{tanC}+{tanC}\:{tanA}=\mathrm{1}. \\ $$ Answered by sandy_suhendra last updated on 04/Oct/16 $$=\mathrm{tanA}\:\mathrm{tanB}\:+\:\mathrm{tanC}\:\left(\mathrm{tanA}+\mathrm{tanB}\right) \\ $$$$=\mathrm{tanA}\:\mathrm{tanB}\:+\:\mathrm{tanC}\:\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right)\left[\mathrm{1}−\mathrm{tanA}\:\mathrm{tanB}\right] \\…
Question Number 8224 by trapti rathaur@ gmail.com last updated on 03/Oct/16 $${if}\:\:\varnothing\:{lies}\:{between}\:\:\:−\frac{\pi}{\mathrm{4}}\:{and}\:\:\frac{\pi}{\mathrm{4}}\:\:\:{then}\:\:{prove}\:{that} \\ $$$$\varnothing^{\mathrm{2}} =\mathrm{tan}\:^{\mathrm{2}} \varnothing\:−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\frac{\mathrm{tan}\:^{\mathrm{4}} \varnothing}{\mathrm{2}}\:+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}\right)\frac{\mathrm{tan}\:^{\mathrm{6}} \varnothing}{\mathrm{3}}\:+−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−{to}\:\infty\:{terms} \\ $$$$ \\ $$ Commented…
Question Number 8176 by lepan last updated on 02/Oct/16 $${Prove}\:{that}\:{cos}\mathrm{2}\theta=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}.{Hence}\:{deduce}\:{that}\: \\ $$$${tan}\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}=\sqrt{\mathrm{2}}−\mathrm{1}. \\ $$$$ \\ $$$$ \\ $$ Answered by prakash jain last…
Question Number 73608 by mathocean1 last updated on 13/Nov/19 $$\:\mathrm{determiner} \\ $$$$\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{exacte}\:\mathrm{de}\:\mathrm{sin}\left(\frac{\mathrm{85}\Pi}{\mathrm{4}}\right) \\ $$$$ \\ $$ Commented by mathmax by abdo last updated on 13/Nov/19…
Question Number 139141 by bramlexs22 last updated on 23/Apr/21 $$\:\:\:\mathrm{3}^{\mathrm{sin}\:\mathrm{x}} \:\mathrm{cos}\:\mathrm{x}\:−\mathrm{3}^{\mathrm{cos}\:\mathrm{x}} \:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{0} \\ $$ Answered by phanphuoc last updated on 23/Apr/21 $$\frac{\mathrm{3}^{{sinx}} }{{sinx}}=\frac{\mathrm{3}^{{cosx}} }{{cosx}}\rightarrow{sinx}={cosx}\rightarrow{tgx}=\mathrm{1}\rightarrow{x}=\pi/\mathrm{4}+{k}\pi \\…
Question Number 73544 by Rio Michael last updated on 13/Nov/19 $${expressf}\left(\theta\right)=\:\mathrm{8}{cos}\theta\:−\mathrm{15}{sin}\theta\:{in}\:{the}\:{form} \\ $$$$\:{rcos}\left(\theta\:+\:\alpha\right),\:{where}\:{r}>\mathrm{0}\:{and}\:\alpha\:{is}\:{a}\:{positive}\:{acute}\:{angle} \\ $$$${hence} \\ $$$${find}\:{the}\:{general}\:{solution}\:{of}\:{the}\:{equation} \\ $$$$\:\:\mathrm{80}{cos}\:\theta\:−\mathrm{150}{sin}\theta\:=\:\mathrm{13} \\ $$$${the}\:{maximum}\:{and}\:{minimum}\:{value}\:{of}\:\:\frac{\mathrm{5}}{{f}\left(\theta\right)\:+\:\mathrm{3}} \\ $$ Answered by…
Question Number 73538 by Rio Michael last updated on 13/Nov/19 $${find}\:{the}\:{general}\:{solution}\:{of}\: \\ $$$$\:{sin}\mathrm{4}{x}\:+\:{cos}\mathrm{2}{x}\:=\:\mathrm{0} \\ $$ Answered by ajfour last updated on 13/Nov/19 $$\mathrm{cos}\:\mathrm{2}{x}\left(\mathrm{2sin}\:\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2}{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\:\:\:{or}…
Question Number 139075 by bramlexs22 last updated on 22/Apr/21 Answered by mr W last updated on 22/Apr/21 $$\mathrm{tan}\:\alpha=\frac{{x}}{\mathrm{144}} \\ $$$$\mathrm{tan}\:\beta=\frac{{x}}{\mathrm{144}+\mathrm{81}}=\frac{{x}}{\mathrm{225}} \\ $$$$\mathrm{tan}\:\gamma=\frac{{x}}{\mathrm{225}+\mathrm{99}}=\frac{{x}}{\mathrm{324}} \\ $$$$\mathrm{tan}\:\gamma=\mathrm{tan}\:\left(\mathrm{90}°−\alpha−\beta\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\beta\right)} \\…
Question Number 7996 by ridwan balatif last updated on 27/Sep/16 $${how}\:{to}\:{prove}\:{this} \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{10}^{{o}} }+\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \mathrm{20}^{{o}} }+\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{40}}=\mathrm{12} \\ $$ Commented by prakash jain last…