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Question Number 12551 by @ANTARES_VY last updated on 25/Apr/17

This y=((2cos^2 x+sin2x)/(2sin^2 x)) find the smallest value of the function.

$$\boldsymbol{\mathrm{This}}\:\:\boldsymbol{\mathrm{y}}=\frac{\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}}{\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\:\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{smallest}}\:\:\boldsymbol{\mathrm{value}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{function}}. \\ $$

Answered by mrW1 last updated on 25/Apr/17

y=((2cos^2 x+sin2x)/(2sin^2 x)) =((2cos^2 x+2sinxcos x)/(2sin^2 x)) =((cos^2 x+sinxcos x)/(sin^2 x)) =cot^2 x+cot x =cot^2 x+2×(1/2)cot x+((1/2))^2 −(1/4) =(cot x+(1/2))^2 −(1/4)≥−(1/4) ⇒smalles value of function=−(1/4) minimum when cot x+(1/2)=0 or tan x=−2 or x=nπ−tan^(−1) (2)

$$\boldsymbol{\mathrm{y}}=\frac{\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}}{\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\: \\ $$$$=\frac{\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{sinx}}\mathrm{cos}\:{x}}{\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\: \\ $$$$=\frac{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{sinx}}\mathrm{cos}\:{x}}{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\: \\ $$$$=\mathrm{cot}^{\mathrm{2}} \:{x}+\mathrm{cot}\:{x} \\ $$$$=\mathrm{cot}^{\mathrm{2}} \:{x}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cot}\:{x}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\left(\mathrm{cot}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\geqslant−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{smalles}\:{value}\:{of}\:{function}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${minimum}\:{when}\:\mathrm{cot}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${or}\:\mathrm{tan}\:{x}=−\mathrm{2} \\ $$$${or}\:{x}={n}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$

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