Question Number 135792 by benjo_mathlover last updated on 16/Mar/21
$${Find}\:{the}\:{minimum}\:{value} \\ $$$${of}\:\mid\mathrm{sin}\:{x}+\mathrm{cos}\:{x}+\mathrm{tan}\:{x}+\mathrm{cot}\:{x}+\mathrm{sec}\:{x}+\mathrm{csc}\:{x}\mid\: \\ $$$${for}\:{real}\:{numbers}\:{x}. \\ $$
Answered by MJS_new last updated on 16/Mar/21
![x=t+((5π)/4) leads to f(t)=((2(√2)cos^3 t +(√2)cos t −2)/(1−2cos^2 t)) this has no real zeros f′(t)=−((4(2(√2)cos^3 t +2cos^2 t −3(√2)cos t +1)sin t)/(((√2)+2cos t)^2 ((√2)−3cos t))) this has zeros sin t =0 cos t =((√2)/2) cos t =1−((√2)/2) [cos t =−1−((√2)/2) rejected] testing all of these we get min ∣f(t)∣ =−1+2(√2)](Q135802.png)
$${x}={t}+\frac{\mathrm{5}\pi}{\mathrm{4}}\:\mathrm{leads}\:\mathrm{to} \\ $$$${f}\left({t}\right)=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{cos}^{\mathrm{3}} \:{t}\:+\sqrt{\mathrm{2}}\mathrm{cos}\:{t}\:−\mathrm{2}}{\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \:{t}} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros} \\ $$$${f}'\left({t}\right)=−\frac{\mathrm{4}\left(\mathrm{2}\sqrt{\mathrm{2}}\mathrm{cos}^{\mathrm{3}} \:{t}\:+\mathrm{2cos}^{\mathrm{2}} \:{t}\:−\mathrm{3}\sqrt{\mathrm{2}}\mathrm{cos}\:{t}\:+\mathrm{1}\right)\mathrm{sin}\:{t}}{\left(\sqrt{\mathrm{2}}+\mathrm{2cos}\:{t}\right)^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{3cos}\:{t}\right)} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{zeros} \\ $$$$\mathrm{sin}\:{t}\:=\mathrm{0} \\ $$$$\mathrm{cos}\:{t}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:{t}\:=\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\left[\mathrm{cos}\:{t}\:=−\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{rejected}\right] \\ $$$$\mathrm{testing}\:\mathrm{all}\:\mathrm{of}\:\mathrm{these}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{min}\:\mid{f}\left({t}\right)\mid\:=−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Answered by liberty last updated on 16/Mar/21
![set { ((sin x=a)),((cos x=b)) :} we want to minimize L=∣a+b+(a/b)+(b/a)+(1/a)+(1/b)∣ = ∣((ab(a+b)+a^2 +b^2 +a+b)/(ab))∣ where a^2 +b^2 =1. let a+b = c we have c^2 =(a+b)^2 =1+2ab so 2ab = c^2 −1. consider c=sin x+cos x=(√2) sin (x+(π/4)) so the range of c is the interval [ −(√2) ,(√2) ]. Consequently it suffices to find the minimum of L(c)=∣((c(c^2 −1)+2(c+1))/(c^2 −1))∣ L(c)=∣c+(2/(c−1))∣, for c in the interval [−(√2) ,(√2) ], taking derivative (dL/dc) = 1−(2/((c−1)^2 )) =0 we get c=1±(√2) . Testing for c=1−(√2) ⇒L=∣1−(√2) +(2/(1−(√2)−1))∣ = ∣1−(√2) −(√2) ∣=2(√2)−1 (min) for c = 1+(√2) →rejected for max value if c = (√2) we get L=∣(√2) +(2/( (√2)−1))∣ = ∣(√2)+2(√2)+2∣=3(√2)+2](Q135815.png)
$${set}\:\begin{cases}{\mathrm{sin}\:{x}={a}}\\{\mathrm{cos}\:{x}={b}}\end{cases} \\ $$$${we}\:{want}\:{to}\:{minimize}\: \\ $$$${L}=\mid{a}+{b}+\frac{{a}}{{b}}+\frac{{b}}{{a}}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\mid \\ $$$$=\:\mid\frac{{ab}\left({a}+{b}\right)+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}+{b}}{{ab}}\mid \\ $$$${where}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}.\:{let}\:{a}+{b}\:=\:{c} \\ $$$${we}\:{have}\:{c}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{ab} \\ $$$${so}\:\mathrm{2}{ab}\:=\:{c}^{\mathrm{2}} −\mathrm{1}.\:{consider}\: \\ $$$${c}=\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right) \\ $$$${so}\:{the}\:{range}\:{of}\:{c}\:{is}\:{the}\:{interval} \\ $$$$\left[\:−\sqrt{\mathrm{2}}\:,\sqrt{\mathrm{2}}\:\right].\:{Consequently} \\ $$$${it}\:{suffices}\:{to}\:{find}\:{the}\:{minimum} \\ $$$${of}\:{L}\left({c}\right)=\mid\frac{{c}\left({c}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2}\left({c}+\mathrm{1}\right)}{{c}^{\mathrm{2}} −\mathrm{1}}\mid \\ $$$${L}\left({c}\right)=\mid{c}+\frac{\mathrm{2}}{{c}−\mathrm{1}}\mid,\:{for}\:{c}\:{in}\:{the} \\ $$$${interval}\:\left[−\sqrt{\mathrm{2}}\:,\sqrt{\mathrm{2}}\:\right],\:{taking} \\ $$$${derivative}\:\frac{{dL}}{{dc}}\:=\:\mathrm{1}−\frac{\mathrm{2}}{\left({c}−\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${we}\:{get}\:{c}=\mathrm{1}\pm\sqrt{\mathrm{2}}\:.\:{Testing}\:{for} \\ $$$${c}=\mathrm{1}−\sqrt{\mathrm{2}}\:\Rightarrow{L}=\mid\mathrm{1}−\sqrt{\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{1}−\sqrt{\mathrm{2}}−\mathrm{1}}\mid \\ $$$$=\:\mid\mathrm{1}−\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{2}}\:\mid=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\:\left({min}\right) \\ $$$${for}\:{c}\:=\:\mathrm{1}+\sqrt{\mathrm{2}}\:\rightarrow{rejected}\: \\ $$$${for}\:{max}\:{value}\:{if}\:{c}\:=\:\sqrt{\mathrm{2}} \\ $$$${we}\:{get}\:{L}=\mid\sqrt{\mathrm{2}}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\mid\:=\:\mid\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\mid=\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$
Commented by liberty last updated on 16/Mar/21
Commented by liberty last updated on 16/Mar/21
