Question Number 42673 by Tawa1 last updated on 31/Aug/18 Answered by OrestesDante last updated on 15/Sep/18 $$\frac{{p}}{{q}}=\frac{{sin}\mathrm{2}\alpha+{sin}\mathrm{2}\beta}{{cos}\mathrm{2}\alpha+{cos}\mathrm{2}\beta} \\ $$$$\frac{\mathrm{2}{sin}\left(\frac{\mathrm{2}\alpha+\mathrm{2}\beta}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{2}\alpha−\mathrm{2}\beta}{\mathrm{2}}\right)}{\mathrm{2}{cos}\left(\frac{\mathrm{2}\alpha+\mathrm{2}\beta}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{2}\alpha−\mathrm{2}\beta}{\mathrm{2}}\right)}=\frac{{p}}{{q}} \\ $$$${tan}\left(\alpha+\beta\right)=\frac{{p}}{{q}} \\ $$$$>>>>>>>>>>>>>>>> \\ $$$$…
Question Number 108145 by bemath last updated on 15/Aug/20 $$\:\:\:\frac{\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{\Pi} \\ $$$$\:\left(\mathrm{1}\right)\:\:\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\:>\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)\: \\ $$$$\:\left(\mathrm{2}\right)\:\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$$$\left(\mathrm{3}\right)\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\sqrt{\mathrm{tan}\:{x}}}{\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$ Commented by…
Question Number 108085 by bemath last updated on 14/Aug/20 $$\:\:\:\:\:\frac{\Cup\mathcal{B}{e}\mathcal{M}{ath}\Cup}{\infty} \\ $$$$\:\:\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{75}°−\mathrm{cos}\:^{\mathrm{8}} \mathrm{75}°\:= \\ $$ Commented by bemath last updated on 14/Aug/20 $${thank}\:{you}\:{both} \\…
Question Number 42520 by gyugfeet last updated on 27/Aug/18 $${cos}^{\mathrm{3}} \:{A}.{sin}\mathrm{3}{A}+{sinA}.{cos}\mathrm{3}{A}=\frac{\mathrm{3}}{\mathrm{4}}{sin}\mathrm{4}{A} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18 $${cos}\mathrm{3}{A}=\frac{\mathrm{4}{cos}^{\mathrm{3}} {A}−\mathrm{3}{cosA}}{} \\ $$$$\left(\frac{\mathrm{3}{cosA}+{cos}\mathrm{3}{A}}{\mathrm{4}}\right){sin}\mathrm{3}{A}+{sinAcos}\mathrm{3}{A} \\…
Question Number 42519 by gyugfeet last updated on 27/Aug/18 $${cosec}\mathrm{2}{A}+{cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}={cotA}−{cot}\mathrm{8}{A}\left({prlve}\:{ghis}\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18 $${cotA}−\left({cosec}\mathrm{2}{A}+{cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$${cotA}−{cose}\mathrm{2}{A}−\left({cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$$\frac{{cosA}}{{sinA}}−\frac{\mathrm{1}}{\mathrm{2}{sinAcosA}}−\left({cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\…
Question Number 107978 by anonymous last updated on 13/Aug/20 $${on}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\pi\right]\:{solve} \\ $$$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{3}{secx}=\:−\mathrm{3} \\ $$$$ \\ $$ Answered by bemath last updated on 14/Aug/20 $$\:\:\:\frac{\bullet\mathscr{B}{emath}\bullet}{\mathscr{C}{ooll}}…
Question Number 107977 by anonymous last updated on 13/Aug/20 $${On}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\mathrm{2}\pi\right]\:{solve} \\ $$$$\mathrm{sin}\:\mathrm{6}{x}\:+\mathrm{sin}\:\mathrm{2}{x}=\mathrm{0} \\ $$$$ \\ $$ Answered by bemath last updated on 13/Aug/20 $$\:\:\frac{\mathcal{B}{e}\frac{\mathcal{M}{ath}}{\heartsuit}}{{joss}} \\…
Question Number 107932 by bemath last updated on 13/Aug/20 $$\:\:\:\frac{\mathbb{B}{e}\mathbb{M}{ath}}{\bullet} \\ $$$${Given}\:\begin{cases}{\mathrm{tan}\:\left({x}−{y}\right)=\frac{\mathrm{3}}{\mathrm{4}}}\\{\mathrm{tan}\:{x}\:=\:\mathrm{2}\:}\end{cases} \\ $$$${find}\:\:\mathrm{tan}\:{y}\:? \\ $$ Commented by mnjuly1970 last updated on 13/Aug/20 $$\:{excellent}. \\…
Question Number 173447 by ajfour last updated on 11/Jul/22 $${Find}\:\mathrm{tan}\:\left(\mathrm{142}.\mathrm{5}°\right)\:{without} \\ $$$${tables}. \\ $$ Commented by mr W last updated on 12/Jul/22 $$\boldsymbol{{welcome}}\:\boldsymbol{{back}}\:\boldsymbol{{sir}}! \\ $$$$…
Question Number 107900 by bemath last updated on 13/Aug/20 $$\:\:\:\:\:\:\:\frac{\Sigma\:\mathcal{B}{e}\mathcal{M}{ath}\:\Sigma}{\Box} \\ $$$$\:\:{Given}\:\mathrm{tan}\:{x}−\mathrm{sec}\:{x}\:=\:\vartheta\: \\ $$$$\:{then}\:\mathrm{sin}\:{x}\:=\:? \\ $$ Answered by bemath last updated on 13/Aug/20 $$\frac{\mathrm{sin}\:{x}−\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\upsilon\:\Rightarrow\:\mathrm{sin}\:{x}−\mathrm{1}=\upsilon\:\mathrm{cos}\:{x}\:…\left(\mathrm{1}\right) \\…