Question Number 12092 by agni5 last updated on 13/Apr/17 $$\mathrm{Prove}\:\mathrm{that}\:\:\mathrm{tan}\:\mathrm{70}°−\mathrm{tan}\:\mathrm{50}°+\mathrm{tan}\:\mathrm{10}°\:=\sqrt{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12085 by chux last updated on 12/Apr/17 Answered by ajfour last updated on 12/Apr/17 $${let}\:\mathrm{6}°=\theta \\ $$$$\mathrm{sin}\:\mathrm{24}°=\mathrm{2sin}\:\mathrm{12}°\mathrm{cos}\:\mathrm{12}° \\ $$$${let}\:\mathrm{sin}\:\mathrm{12}°={x}\:\:{and}\:\:\mathrm{cos}\:\mathrm{12}°={y} \\ $$$$\mathrm{sin}\:\left(\mathrm{3}\theta+\mathrm{2}\theta\right)=\:{y}\:\mathrm{sin}\:\mathrm{18}°+{x}\:\mathrm{cos}\:\mathrm{18}° \\ $$$$\Rightarrow\:{x}\mathrm{cos}\:\mathrm{18}°+{y}\mathrm{sin}\:\mathrm{18}°\:=\mathrm{sin}\:\mathrm{5}\theta=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 12078 by agni5 last updated on 12/Apr/17 $$\mathrm{If}\:\mathrm{tan}\:^{\mathrm{2}} \alpha\:=\:\mathrm{1}+\mathrm{2tan}\:^{\mathrm{2}} \beta\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\mathrm{cos}\:\mathrm{2}\beta\:=\mathrm{1}+\mathrm{2cos}\:\mathrm{2}\alpha\:. \\ $$ Answered by ajfour last updated on 12/Apr/17 $$\mathrm{1}+\mathrm{2cos}\:\mathrm{2}\alpha\:=\mathrm{1}+\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \alpha\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}}…
Question Number 12063 by tawa last updated on 10/Apr/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Apr/17 $$\mathrm{100}+\mathrm{4}\left(\mathrm{180}−{a}\right)=\mathrm{360}\Rightarrow\mathrm{4}{a}=\mathrm{820}−\mathrm{360} \\ $$$$\mathrm{4}{a}=\mathrm{460}\Rightarrow{a}=\mathrm{115}\:\:\:\:.\blacksquare \\ $$ Terms of Service Privacy…
Question Number 143102 by liberty last updated on 10/Jun/21 Answered by som(math1967) last updated on 10/Jun/21 $$\mathrm{2}\boldsymbol{{y}}=\mathrm{40}°+\mathrm{2}×\mathrm{48} \\ $$$$\boldsymbol{{y}}=\frac{\mathrm{136}}{\mathrm{2}}=\mathrm{68}° \\ $$$$\boldsymbol{{again}}\:\mathrm{48}\:+\boldsymbol{{x}}=\boldsymbol{{y}} \\ $$$$\boldsymbol{{x}}=\mathrm{68}−\mathrm{48}=\mathrm{20}° \\ $$…
Question Number 12028 by tawa last updated on 09/Apr/17 $$\mathrm{Express}\::\:\mathrm{sin}\left(\mathrm{33}\right)\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form}. \\ $$ Answered by sandy_suhendra last updated on 10/Apr/17 $$\mathrm{let}\:\mathrm{A}=\mathrm{18}° \\ $$$$\mathrm{5A}=\mathrm{90}° \\ $$$$\mathrm{3A}=\mathrm{90}°−\mathrm{2A} \\…
Question Number 143063 by ERA last updated on 09/Jun/21 $$\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2n}+\mathrm{1}}\right)…..\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\mathrm{n}\pi}}{\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{\mathrm{n}}} } \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143057 by ERA last updated on 09/Jun/21 $$\mathrm{cos}\left(\boldsymbol{\alpha}\right)×\mathrm{cos}\left(\mathrm{2}\alpha\right)×\mathrm{cos}\left(\mathrm{4}\alpha\right)×….×\mathrm{cos}\left(\mathrm{2}^{\mathrm{n}} \boldsymbol{\alpha}\right)=\frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{2}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \boldsymbol{\alpha}\right)}{\mathrm{2}^{\mathrm{n}+\mathrm{1}} \mathrm{sin}\left(\alpha\right)} \\ $$$$\boldsymbol{\mathrm{prove}} \\ $$ Answered by Dwaipayan Shikari last updated on 09/Jun/21…
Question Number 143039 by bramlexs22 last updated on 09/Jun/21 $$\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${x}=? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 77472 by mathocean1 last updated on 06/Jan/20 $$\mathrm{Hello}\:\mathrm{sirs}…\: \\ $$$$\mathrm{i}\:\mathrm{want}\:\mathrm{that}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{me}\:\mathrm{how}\: \\ $$$$\:\mathrm{we}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{trigonometric} \\ $$$$\mathrm{inequality}\:\mathrm{with}\:\mathrm{tangente}.\: \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{take}\:\mathrm{for}\:\mathrm{example}\:\mathrm{tan2x}\geqslant\sqrt{\mathrm{3}} \\ $$$$\mathrm{i}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{type}\:\mathrm{of}\:\mathrm{equality} \\ $$$$\mathrm{but}\:\mathrm{not}\:\mathrm{the}\:\mathrm{inequality}!\:\mathrm{Please}\: \\ $$$$\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help}…\:\mathrm{Even}\:\mathrm{the}\:\mathrm{steps} \\…