Question Number 115328 by bobhans last updated on 25/Sep/20 $${If}\:\frac{\mathrm{sin}\:\mathrm{1}°+\mathrm{sin}\:\mathrm{2}°+\mathrm{sin}\:\mathrm{3}°+…+\mathrm{sin}\:\mathrm{44}°}{\mathrm{cos}\:\mathrm{1}°+\mathrm{cos}\:\mathrm{2}°+\mathrm{cos}\:\mathrm{3}°+…+\mathrm{cos}\:\mathrm{44}°}=\chi \\ $$$${then}\:\chi^{\mathrm{4}} +\mathrm{4}\chi^{\mathrm{3}} +\mathrm{4}\chi^{\mathrm{2}} +\mathrm{4}= \\ $$ Answered by bemath last updated on 25/Sep/20 $$\:\:\mathrm{sin}\:\mathrm{44}°+\mathrm{sin}\:\mathrm{1}°=\mathrm{2sin}\:\left(\frac{\mathrm{45}°}{\mathrm{2}}\right).\mathrm{cos}\:\left(\frac{\mathrm{43}°}{\mathrm{2}}\right)…
Question Number 180826 by mnjuly1970 last updated on 17/Nov/22 $$ \\ $$$$\:\:\:\:\mathrm{I}{f}\:\:,\:\:\:\mathrm{2}{sin}\left(\theta\:\right)−\mathrm{3}{cos}\left(\theta\right)\:=\mathrm{3} \\ $$$$\:\Rightarrow\:\:\:\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\:−\:\mathrm{3}{cos}\left(\frac{\theta}{\mathrm{2}}\right)\:=\:? \\ $$$$ \\ $$ Answered by mr W last updated on…
Question Number 115255 by aye48 last updated on 24/Sep/20 $$\mathrm{Express}\:\mathrm{cosec}\:\mathrm{3x}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{cosec}\:\mathrm{x}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115253 by aye48 last updated on 24/Sep/20 $$\mathrm{Express}\:\mathrm{sin}\:\mathrm{4x}\:\mathrm{interm}\:\mathrm{of}\:\mathrm{sin}\:\mathrm{x}. \\ $$ Answered by Dwaipayan Shikari last updated on 24/Sep/20 $$\mathrm{sin4x}=\mathrm{2sin2xcos2x} \\ $$$$=\mathrm{2sinxcosx}\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\right) \\…
Question Number 115248 by aye48 last updated on 24/Sep/20 $$\mathrm{Express}\:\mathrm{cos}\:\mathrm{4x}\:\mathrm{interm}\:\mathrm{of}\:\mathrm{cos}\:\mathrm{x}. \\ $$ Answered by Olaf last updated on 24/Sep/20 $$\mathrm{cos4}{x}\:=\:\mathrm{2cos}^{\mathrm{2}} \mathrm{2}{x}−\mathrm{1} \\ $$$$\mathrm{cos4}{x}\:=\:\mathrm{2}\left(\mathrm{2cos}^{\mathrm{2}} {x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}…
Question Number 49660 by Rio Michael last updated on 08/Dec/18 Commented by Rio Michael last updated on 08/Dec/18 $${Above}\:{is}\:{a}\:{Circle}\:{with}\:{centre}\:{O}\:.{If}\:{it}\:{has}\:{radius}\:{of}\:\mathrm{10}{cm} \\ $$$${find}\:{the}\:{lenght}\:{of}\:{the}\:{line}\:{AC}.{Note}\::\:\angle{AOC}=\mathrm{30}° \\ $$ Answered by…
Question Number 49605 by Rio Michael last updated on 08/Dec/18 Commented by Rio Michael last updated on 08/Dec/18 $${The}\:{figure}\:{above}\:{has}\:\bigtriangleup{ABC}\:{and}\:\bigtriangleup{MBD}\:{where}\:{AC}\:{is}\:{parallel} \\ $$$${to}\:{DM}.{Find}\:{the}\:{lenght}\:{AM}\:{if} \\ $$$$\left.{a}\right)\:{BM}\:=\:\mathrm{6}.\mathrm{5}{m} \\ $$$$\left.{b}\right)\:{Given}\:{the}\:{the}\:{area}\:{of}\:\bigtriangleup{ABC}\:=\:\mathrm{120}{m}^{\mathrm{2}}…
Question Number 115121 by bemath last updated on 23/Sep/20 $${solve}\:\begin{cases}{\mathrm{tan}\:{x}\:+\:\mathrm{cot}\:{x}\:=\:{p}}\\{\mathrm{sec}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:{q}}\end{cases} \\ $$ Commented by ATHISHHUZAIN last updated on 23/Sep/20 $${f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2}{x}+\mathrm{1}\:\:\:\:\:{o}}\\{\mathrm{sec}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:{q}}\end{cases} \\ $$ Commented by bemath…
Question Number 49534 by aseerimad last updated on 07/Dec/18 Commented by aseerimad last updated on 07/Dec/18 $${how}\:{is}\:{this}\:{done}\:{in}\:{short}?\:{Thanks}\:{in}\:{advance}. \\ $$ Commented by Abdo msup. last updated…
Question Number 115022 by dw last updated on 23/Sep/20 $$\left[\boldsymbol{{Q}}.\mathrm{1}\:\right]\:\:\:{Find}\:{the}\:{domain}\:{of} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\frac{\lfloor{cos}^{−\mathrm{1}} \left({x}^{\mathrm{4}} \right)\rfloor+\mid\lfloor{x}−\mathrm{2}{tan}^{−\mathrm{1}} \left({x}\right)\rfloor\mid+\sqrt{{sin}\left({lnx}\right)}}{\left\{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{7}\right\}+{a}^{\sqrt{{sin}\left({x}\right)+\mathrm{3}{cos}\left({x}\right)}} +{ln}\:{cos}\left(\frac{\mathrm{1}}{\:\sqrt{−{x}^{\mathrm{2}} }}\right)} \\ $$$${where}\:\left\{{x}\right\}\:{reprents}\:\:{the}\:{fractionare}\:{part}\:{of}\:{x}: \\ $$$$ \\…