Question Number 116520 by kaivan.ahmadi last updated on 04/Oct/20 $${find}\:{the}\:{center}\:{of}\:{circle}\:{with}\:{radius}\:{R} \\ $$$${and}\:{tangent}\:{by}\:{sides}\:{AB}\:{and}\:{BC}\:{of}\: \\ $$$${triangle}? \\ $$ Answered by $@y@m last updated on 04/Oct/20 $${The}\:{centre}\:{of}\:{the}\:{circle}\:{is}\:{Incentre}\:{of} \\…
Question Number 116503 by bemath last updated on 04/Oct/20 $$\mathrm{If}\:\mathrm{tan}\:\alpha\:\mathrm{and}\:\mathrm{tan}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\mathrm{equation}\:\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{find}\:\begin{cases}{\mathrm{tan}\:\left(\frac{\alpha}{\mathrm{2}}+\mathrm{2}\beta\right)}\\{\mathrm{tan}\:\left(\mathrm{2}\alpha−\frac{\beta}{\mathrm{2}}\right)}\end{cases} \\ $$ Answered by bobhans last updated on 04/Oct/20 $$\Rightarrow\mathrm{x}_{\mathrm{1}}…
Question Number 116493 by bemath last updated on 04/Oct/20 $$\mathrm{Given}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{1}\:,\:\mathrm{find}\: \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}\:^{\mathrm{8}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\:=? \\ $$ Answered by bobhans last updated on 04/Oct/20…
Question Number 116483 by bemath last updated on 04/Oct/20 $$\mathrm{If}\:\mathrm{8}\:\mathrm{cos}\:\mathrm{x}−\mathrm{8}\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{3}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{55}\:\mathrm{tan}\:\mathrm{x}\:+\:\frac{\mathrm{55}}{\mathrm{tan}\:\mathrm{x}}\:? \\ $$ Answered by mr W last updated on 04/Oct/20 $$\mathrm{8}\left(\mathrm{1}−\mathrm{tan}\:{x}\right)=\frac{\mathrm{3}}{\mathrm{cos}\:{x}} \\ $$$$\mathrm{64}\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\mathrm{2}}…
Question Number 116448 by bobhans last updated on 04/Oct/20 $$\mathrm{If}\:\mathrm{sin}\:\left(\mathrm{P}\right)\:=\:−\frac{\mathrm{8}}{\mathrm{17}}\:,\mathrm{where}\:\frac{\mathrm{3}\pi}{\mathrm{2}}<\mathrm{P}<\mathrm{2}\pi \\ $$$$\mathrm{and}\:\mathrm{cos}\:\left(\mathrm{Q}\right)=−\frac{\mathrm{24}}{\mathrm{25}},\:\mathrm{where}\:\pi<\mathrm{Q}<\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}\:\left(\mathrm{2P}+\mathrm{Q}\right)\:=? \\ $$ Answered by bemath last updated on 04/Oct/20 $$\left(\mathrm{1}\right)\:\mathrm{cos}\:\left(\mathrm{P}\right)=\:\frac{\mathrm{15}}{\mathrm{17}}\:;\:\mathrm{sin}\:\left(\mathrm{Q}\right)=−\frac{\mathrm{7}}{\mathrm{25}} \\…
Question Number 116441 by bobhans last updated on 04/Oct/20 $$\mathrm{If}\:\mathrm{sin}\:\left(\mathrm{45}°−\mathrm{x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\:,\:\mathrm{where}\:\mathrm{45}°<\mathrm{x}<\mathrm{90}° \\ $$$$.\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{6sin}\:\mathrm{x}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$ Answered by bemath last updated on 04/Oct/20 $$\Rightarrow\mathrm{sin}\:\left(\mathrm{x}−\mathrm{45}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\…
Question Number 116374 by bemath last updated on 03/Oct/20 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\mathrm{2}}\:\mathrm{where}\:\mathrm{x}\:\mathrm{ranges}\:\mathrm{over}\:\mathrm{all} \\ $$$$\mathrm{real}\:\mathrm{numbers}. \\ $$ Answered by MJS_new last updated on 03/Oct/20 $${x}=\mathrm{2arctan}\:{t} \\…
Question Number 181863 by greougoury555 last updated on 01/Dec/22 $$\:\mathrm{cos}\:\mathrm{56}°.\:\mathrm{cos}\:\left(\mathrm{2}.\mathrm{56}°\right).\:\mathrm{cos}\:\left(\mathrm{2}^{\mathrm{2}} .\mathrm{56}°\right).\:\mathrm{cos}\:\left(\mathrm{2}^{\mathrm{3}} .\mathrm{56}°\right)…\:\mathrm{cos}\:\left(\mathrm{2}^{\mathrm{23}} .\mathrm{56}°\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116303 by Engr_Jidda last updated on 02/Oct/20 $${Solve}\:{for}\:{x}\:{in}\:\varrho^{{x}} +{x}=\mathrm{4} \\ $$ Answered by mr W last updated on 02/Oct/20 $$\varrho^{{x}} =\mathrm{4}−{x} \\ $$$$\varrho^{\mathrm{4}}…
Question Number 116290 by ravisoni last updated on 02/Oct/20 Commented by Dwaipayan Shikari last updated on 03/Oct/20 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}^{−\mathrm{1}} \mathrm{x} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}}…