Question Number 157925 by cortano last updated on 29/Oct/21 Answered by FongXD last updated on 30/Oct/21 $$\mathrm{Given}:\:\frac{\mathrm{sin}\alpha\mathrm{cos}\alpha}{\mathrm{sin}\beta\mathrm{cos}\beta}=\frac{\mathrm{8}}{\mathrm{5}}\:\:\:\left(\mathrm{1}\right)\:\mathrm{and}\:\frac{\mathrm{sin}\alpha\mathrm{cos}\beta}{\mathrm{sin}\beta\mathrm{cos}\alpha}=\mathrm{4}\:\:\:\left(\mathrm{2}\right) \\ $$$$\Leftrightarrow\:\left(\mathrm{1}\right)\centerdot\left(\mathrm{2}\right):\:\frac{\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{sin}^{\mathrm{2}} \beta}=\frac{\mathrm{32}}{\mathrm{5}}\:\mathrm{and}\:\left(\mathrm{1}\right)\boldsymbol{\div}\left(\mathrm{2}\right):\:\frac{\mathrm{cos}^{\mathrm{2}} \alpha}{\mathrm{cos}^{\mathrm{2}} \beta}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Leftrightarrow\:\mathrm{5}−\mathrm{5sin}^{\mathrm{2}}…
Question Number 26778 by tawa tawa last updated on 29/Dec/17 $$\mathrm{If}\:\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{A}\right)\:+\:\mathrm{sin}\left(\mathrm{B}\right)\:=\:\mathrm{p}\:\:\:\:\:\mathrm{and}\:\:\:\:\mathrm{cos}\left(\mathrm{A}\right)\:+\:\mathrm{cos}\left(\mathrm{B}\right)\:=\:\mathrm{q} \\ $$$$\mathrm{show}\:\mathrm{that}\:\:\:\:\mathrm{tan}\left(\mathrm{A}\right)\:+\:\mathrm{tan}\left(\mathrm{B}\right)\:=\:\frac{\mathrm{8pq}}{\left(\mathrm{p}^{\mathrm{2}} \:+\:\mathrm{q}^{\mathrm{2}} \right)^{\mathrm{2}} \:−\:\mathrm{4q}^{\mathrm{2}} } \\ $$ Commented by tawa tawa last updated…
Question Number 26713 by Bousselham2 last updated on 28/Dec/17 $$\mathrm{2sin}\:\left(\mathrm{3}{x}\right)<\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157774 by mba last updated on 27/Oct/21 $$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 26561 by tawa tawa last updated on 26/Dec/17 $$\mathrm{Express}\:\:\mathrm{sin}\left(\mathrm{20}\right)°\:\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 92054 by john santu last updated on 04/May/20 $$\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157452 by tounghoungko last updated on 23/Oct/21 $$\:{solve}\:\mathrm{tan}\:{x}=\mathrm{cos}\:\mathrm{3}{x}\: \\ $$ Commented by MJS_new last updated on 23/Oct/21 $$\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{possible}\:\Rightarrow\:\mathrm{you} \\ $$$$\mathrm{must}\:\mathrm{approximate} \\ $$ Terms…
Question Number 157324 by ZiYangLee last updated on 22/Oct/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}\:\mathrm{7}{x}\:−\:\mathrm{cos}\:\mathrm{4}{x}\:+\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$ Commented by cortano last updated on 22/Oct/21 $$\:\mathrm{cos}\:\mathrm{7}{x}+\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0} \\ $$$$\:\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0} \\…
Question Number 157315 by amin96 last updated on 22/Oct/21 Answered by gsk2684 last updated on 22/Oct/21 $$\underset{{t}=\mathrm{5555}} {\overset{\mathrm{55555}} {\prod}}\left(−\mathrm{1}\right)^{{t}} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{5555}} \left(−\mathrm{1}\right)^{\mathrm{5556}} …\left(−\mathrm{1}\right)^{\mathrm{55555}} \\ $$$$=\left(−\mathrm{1}\right)\left(\mathrm{1}\right)\left(−\mathrm{1}\right)….\left(−\mathrm{1}\right)\:\left({odd}\:{number}\:{of}\:{terms}\right)…
Question Number 157227 by amin96 last updated on 21/Oct/21 $$\boldsymbol{\mathrm{x}}\left(\mathrm{3}\boldsymbol{\mathrm{sin}}\left(\sqrt{\boldsymbol{\mathrm{x}}}\right)−\mathrm{2}\sqrt{\boldsymbol{\mathrm{x}}}\right)=\boldsymbol{\mathrm{sin}}^{\mathrm{3}} \left(\sqrt{\boldsymbol{\mathrm{x}}}\right) \\ $$ Answered by Javokhir last updated on 21/Oct/21 $$ \\ $$$$\:\:\:\:\:\:\:\sqrt{{x}}={t} \\ $$$$\:\:\:\:\:\:\:{t}^{\mathrm{2}}…