Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 153736 by ZiYangLee last updated on 09/Sep/21

Given that 7 cos 2θ+24 sin^2 θ=R cos(2θ−α),  where R>0 and 0<α<(π/2), find the maximum  value of 14 cos^2 θ+48 sin θ cos θ.

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{7}\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{24}\:\mathrm{sin}^{\mathrm{2}} \theta={R}\:\mathrm{cos}\left(\mathrm{2}\theta−\alpha\right), \\ $$ $$\mathrm{where}\:{R}>\mathrm{0}\:\mathrm{and}\:\mathrm{0}<\alpha<\frac{\pi}{\mathrm{2}},\:\mathrm{find}\:\mathrm{the}\:\mathrm{maximum} \\ $$ $$\mathrm{value}\:\mathrm{of}\:\mathrm{14}\:\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{48}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta. \\ $$

Answered by mr W last updated on 09/Sep/21

7 cos 2θ+24 sin^2 θ=R cos(2θ−α)  7 cos 2θ−12cos 2θ+12=R cos(2θ−α)  −5cos 2θ+12=R cos2θ cos α−R sin 2θ sin α  12=(R cos α+5) cos2θ−R sin α sin 2θ  ((12)/( (√((Rsin α)^2 +(R cos α+5)^2 ))))=cos ϕ cos2θ−sin ϕ sin 2θ  ((12)/( (√(R^2 +25+10R cos α))))=cos(2θ+ϕ)  2θ=cos^(−1) ((12)/( (√(R^2 +25+10R cos α))))−ϕ  cos 2θ=((12(R cos α+5))/( R^2 +25+10R cos α))+((R sin α(√(R^2 −119+10R cos α)))/( R^2 +25+10R cos α))  sin 2θ=(((R cos α+5)(√(R^2 −119+10R cos α)))/( R^2 +25+10R cos α))−((12R sin α)/( R^2 +25+10R cos α))    14 cos^2 θ+48 sin θ cos θ  =7 cos 2θ+24 sin 2θ+7  =((7(12(R cos α+5)+R sin α(√(R^2 −119+10R cos α)))+24((R cos α+5)(√(R^2 −119+10R cos α))−12R sin α))/(R^2 +25+10R cos α))+7  =((84R cos α−288R sin α+420+(24R cos α+7R sin α+120)(√(R^2 −119+10R cos α)))/(R^2 +25+10R cos α))+7  .....

$$\mathrm{7}\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{24}\:\mathrm{sin}^{\mathrm{2}} \theta={R}\:\mathrm{cos}\left(\mathrm{2}\theta−\alpha\right) \\ $$ $$\mathrm{7}\:\mathrm{cos}\:\mathrm{2}\theta−\mathrm{12cos}\:\mathrm{2}\theta+\mathrm{12}={R}\:\mathrm{cos}\left(\mathrm{2}\theta−\alpha\right) \\ $$ $$−\mathrm{5cos}\:\mathrm{2}\theta+\mathrm{12}={R}\:\mathrm{cos2}\theta\:\mathrm{cos}\:\alpha−{R}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{sin}\:\alpha \\ $$ $$\mathrm{12}=\left({R}\:\mathrm{cos}\:\alpha+\mathrm{5}\right)\:\mathrm{cos2}\theta−{R}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\mathrm{2}\theta \\ $$ $$\frac{\mathrm{12}}{\:\sqrt{\left({R}\mathrm{sin}\:\alpha\right)^{\mathrm{2}} +\left({R}\:\mathrm{cos}\:\alpha+\mathrm{5}\right)^{\mathrm{2}} }}=\mathrm{cos}\:\varphi\:\mathrm{cos2}\theta−\mathrm{sin}\:\varphi\:\mathrm{sin}\:\mathrm{2}\theta \\ $$ $$\frac{\mathrm{12}}{\:\sqrt{{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}}=\mathrm{cos}\left(\mathrm{2}\theta+\varphi\right) \\ $$ $$\mathrm{2}\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{12}}{\:\sqrt{{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}}−\varphi \\ $$ $$\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{12}\left({R}\:\mathrm{cos}\:\alpha+\mathrm{5}\right)}{\:{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}+\frac{{R}\:\mathrm{sin}\:\alpha\sqrt{{R}^{\mathrm{2}} −\mathrm{119}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}}{\:{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha} \\ $$ $$\mathrm{sin}\:\mathrm{2}\theta=\frac{\left({R}\:\mathrm{cos}\:\alpha+\mathrm{5}\right)\sqrt{{R}^{\mathrm{2}} −\mathrm{119}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}}{\:{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}−\frac{\mathrm{12}{R}\:\mathrm{sin}\:\alpha}{\:{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha} \\ $$ $$ \\ $$ $$\mathrm{14}\:\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{48}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$ $$=\mathrm{7}\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{24}\:\mathrm{sin}\:\mathrm{2}\theta+\mathrm{7} \\ $$ $$=\frac{\mathrm{7}\left(\mathrm{12}\left({R}\:\mathrm{cos}\:\alpha+\mathrm{5}\right)+{R}\:\mathrm{sin}\:\alpha\sqrt{{R}^{\mathrm{2}} −\mathrm{119}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}\right)+\mathrm{24}\left(\left({R}\:\mathrm{cos}\:\alpha+\mathrm{5}\right)\sqrt{{R}^{\mathrm{2}} −\mathrm{119}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}−\mathrm{12}{R}\:\mathrm{sin}\:\alpha\right)}{{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}+\mathrm{7} \\ $$ $$=\frac{\mathrm{84}{R}\:\mathrm{cos}\:\alpha−\mathrm{288}{R}\:\mathrm{sin}\:\alpha+\mathrm{420}+\left(\mathrm{24}{R}\:\mathrm{cos}\:\alpha+\mathrm{7}{R}\:\mathrm{sin}\:\alpha+\mathrm{120}\right)\sqrt{{R}^{\mathrm{2}} −\mathrm{119}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}}{{R}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{R}\:\mathrm{cos}\:\alpha}+\mathrm{7} \\ $$ $$..... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com