Question Number 96758 by bobhans last updated on 04/Jun/20
![Let x∈ [ −((5π)/(12)) , −(π/3) ] . The maximum value of y = tan (x+((2π)/3))−tan (x+(π/6)) +cos (x+(π/6)) is ___](Q96758.png)
$$\mathrm{Let}\:{x}\in\:\left[\:−\frac{\mathrm{5}\pi}{\mathrm{12}}\:,\:−\frac{\pi}{\mathrm{3}}\:\right]\:.\:\mathrm{The}\:\mathrm{maximum}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{y}\:=\:\mathrm{tan}\:\left({x}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{6}}\right)\:+\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{6}}\right) \\ $$$$\mathrm{is}\:\_\_\_ \\ $$
Commented by john santu last updated on 04/Jun/20
![set m = −x−(π/6), m∈ [ (π/6), (π/4) ] and 2m ∈ [ (π/3), (π/2) ]. we have tan (x+((2π)/3)) = −cot (x+(π/6)) = cot m then y = cot m + tan m + cos m y = (2/(sin 2m)) + cos m. since both (2/(sin 2m)) and cos m are monotonic decreasing in this case, so y reaches the maksimum at m = (π/6) , where y_(max) = (2/(sin (π/3))) + cos (π/6) = (4/((√3) )) + ((√3)/2) = ((11(√3))/6) .](Q96759.png)
$$\mathrm{set}\:{m}\:=\:−{x}−\frac{\pi}{\mathrm{6}},\:{m}\in\:\left[\:\frac{\pi}{\mathrm{6}},\:\frac{\pi}{\mathrm{4}}\:\right] \\ $$$$\mathrm{and}\:\mathrm{2}{m}\:\in\:\left[\:\frac{\pi}{\mathrm{3}},\:\frac{\pi}{\mathrm{2}}\:\right].\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\mathrm{tan}\:\left({x}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:=\:−\mathrm{cot}\:\left({x}+\frac{\pi}{\mathrm{6}}\right)\:=\:\mathrm{cot}\:{m} \\ $$$$\mathrm{then}\:{y}\:=\:\mathrm{cot}\:{m}\:+\:\mathrm{tan}\:{m}\:+\:\mathrm{cos}\:{m} \\ $$$${y}\:=\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}{m}}\:+\:\mathrm{cos}\:{m}.\: \\ $$$${since}\:{both}\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2}{m}}\:\mathrm{and}\:\mathrm{cos}\:{m}\:{are} \\ $$$${monotonic}\:{decreasing}\:{in}\:{this} \\ $$$${case},\:\mathrm{so}\:\mathrm{y}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{maksimum} \\ $$$$\mathrm{at}\:{m}\:=\:\frac{\pi}{\mathrm{6}}\:,\:\mathrm{where}\:\mathrm{y}_{\mathrm{max}} \:=\:\frac{\mathrm{2}}{\mathrm{sin}\:\frac{\pi}{\mathrm{3}}}\:+\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}} \\ $$$$=\:\frac{\mathrm{4}}{\sqrt{\mathrm{3}}\:}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\frac{\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{6}}\:.\: \\ $$