Menu Close

h-x-sin-4-x-cos-4-x-2msinxcosx-Find-all-the-values-of-the-parameter-m-for-the-funtion-denined-on-R-

Tinku Tara 0 Comments
Pin




Question Number 41911 by lucha116 last updated on 15/Aug/18
h(x)=(√(sin^4 x+cos^4 x−2msinxcosx)) Find all the values of the parameter m for the funtion denined on R
$${h}\left({x}\right)=\sqrt{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}−\mathrm{2}{msinxcosx}} \\ $$$${Find}\:{all}\:{the}\:{values}\:{of}\:{the}\:{parameter}\:{m}\:{for}\:{the}\:{funtion}\:{denined}\:{on}\:{R} \\ $$$$ \\ $$
Commented by MJS last updated on 15/Aug/18
draw the function for ∣m∣>(1/2), you will see that it′s not defined ∀x∈R m=1 ⇒ sin^4 (π/4) +cos^4 (π/4) −2sin (π/4) cos (π/4)= =(1/4)+(1/4)−2×((√2)/2)×((√2)/2)=−(1/2) as I hopefully showed the minimum of sin^4 x +cos^4 x −2msin x cos x must be ≥0 and it′s exactly zero for m=±(1/2). for ∣m∣>(1/2) the minimum is below zero
$$\mathrm{draw}\:\mathrm{the}\:\mathrm{function}\:\mathrm{for}\:\mid{m}\mid>\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{you}\:\mathrm{will}\:\mathrm{see} \\ $$$$\mathrm{that}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{defined}\:\forall{x}\in\mathbb{R} \\ $$$${m}=\mathrm{1}\:\Rightarrow\:\mathrm{sin}^{\mathrm{4}} \:\frac{\pi}{\mathrm{4}}\:+\mathrm{cos}^{\mathrm{4}} \:\frac{\pi}{\mathrm{4}}\:−\mathrm{2sin}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{as}\:\mathrm{I}\:\mathrm{hopefully}\:\mathrm{showed}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x}\:−\mathrm{2}{m}\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:\mathrm{must}\:\mathrm{be}\:\geqslant\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{exactly}\:\mathrm{zero}\:\mathrm{for}\:{m}=\pm\frac{\mathrm{1}}{\mathrm{2}}.\:\mathrm{for}\:\mid{m}\mid>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{below}\:\mathrm{zero} \\ $$
Commented by math khazana by abdo last updated on 15/Aug/18
we have h(x)=(√((cos^2 x+sin^2 x)^2 −2cos^2 xsin^2 x−2msinxcosx)) =(√(1−(1/2)sin^2 (2x)−msin(2x))) =(1/( (√2)))(√(2−sin^2 (2x)−2msin(2x))) due to D_f we mist have −sin^2 (2x)−2msin(2x)+2≥0 ⇒ sin^2 (2x) +2msin(2x)−2≤0 let sin(2x) =t ⇒ t^2 +2mt −2≤0 and ∣t∣≤1 ⇒ Δ^′ =m^2 +2>0⇒t_1 =−m+(√(m^2 +2)) and t_2 =−m−(√(m^(2 ) +2)) ∣t_1 ∣ ≤1 ⇒ −1≤−m+(√(m^2 +2))≤1 ⇒ m−1 ≤(√(m^2 +2))≤m+1 if m>1 ⇒ m^2 −2m+1 ≤m^2 +2 ≤m^2 +2m +2 ⇒ −2m+1≤2≤2m+2 ⇒ −m+(1/2)≤ 1≤ m+1 ⇒ −m≤(1/2) and m≥0 ⇒m≥0⇒ m≥1 if −1<m<1 ⇒−(1−m)≤(√(m^2 +2))≤m+1 ⇒ m^2 −2m +1 ≤m^2 +2 ≤m^2 +2m+1 ⇒ −m≤(1/2)and m≥0 ⇒m≥0 ⇒ 0≤m≤1 ∣t_2 ∣≤1 ⇒ ∣−m−(√(m^2 +2))∣≤1 ⇒ −1≤m+(√(m^2 +2))≤1 ⇒−m−1≤(√(m^2 +2))≤1−m so m≤1 if −m−1>0 ⇒m≤−1 ⇒ m^2 +2m+1≤m^2 +2≤m^2 −2m +1 ⇒ 2m+1≤2≤−2m+1 ⇒ 2m≤1≤−2m ⇒ m≤(1/2) and (1/2)≤−m ⇒m≤(1/2) and m≤−(1/2) ⇒ m≤−(1/2) ⇒m ≤−1 due to initial condition if −m−1≤0 ⇒−1≤m≤1 so ∣t_2 ∣ ≤1 ⇒ −1≤m≤1 finally ∣t_1 ∣≤1 and ∣t_2 ∣≤1 ⇒ 0≤m≤1 .
$${we}\:{have}\:{h}\left({x}\right)=\sqrt{\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{cos}^{\mathrm{2}} {xsin}^{\mathrm{2}} {x}−\mathrm{2}{msinxcosx}} \\ $$$$=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−{msin}\left(\mathrm{2}{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\mathrm{2}−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{2}{msin}\left(\mathrm{2}{x}\right)}\:{due}\:{to}\:{D}_{{f}} {we}\:{mist} \\ $$$${have}\:−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{2}{msin}\left(\mathrm{2}{x}\right)+\mathrm{2}\geqslant\mathrm{0}\:\Rightarrow \\ $$$${sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:+\mathrm{2}{msin}\left(\mathrm{2}{x}\right)−\mathrm{2}\leqslant\mathrm{0}\:{let}\:{sin}\left(\mathrm{2}{x}\right)\:={t}\:\Rightarrow \\ $$$${t}^{\mathrm{2}} +\mathrm{2}{mt}\:−\mathrm{2}\leqslant\mathrm{0}\:\:{and}\:\:\:\mid{t}\mid\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\Delta^{'} \:={m}^{\mathrm{2}} \:+\mathrm{2}>\mathrm{0}\Rightarrow{t}_{\mathrm{1}} =−{m}+\sqrt{{m}^{\mathrm{2}} +\mathrm{2}}\:\:{and}\: \\ $$$${t}_{\mathrm{2}} =−{m}−\sqrt{{m}^{\mathrm{2}\:} +\mathrm{2}}\:\: \\ $$$$\mid{t}_{\mathrm{1}} \mid\:\leqslant\mathrm{1}\:\Rightarrow\:−\mathrm{1}\leqslant−{m}+\sqrt{{m}^{\mathrm{2}} \:+\mathrm{2}}\leqslant\mathrm{1}\:\Rightarrow \\ $$$${m}−\mathrm{1}\:\leqslant\sqrt{{m}^{\mathrm{2}} +\mathrm{2}}\leqslant{m}+\mathrm{1}\:\:\:{if}\:{m}>\mathrm{1}\:\Rightarrow \\ $$$${m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{1}\:\leqslant{m}^{\mathrm{2}} \:+\mathrm{2}\:\leqslant{m}^{\mathrm{2}} \:+\mathrm{2}{m}\:+\mathrm{2}\:\Rightarrow \\ $$$$−\mathrm{2}{m}+\mathrm{1}\leqslant\mathrm{2}\leqslant\mathrm{2}{m}+\mathrm{2}\:\Rightarrow\:−{m}+\frac{\mathrm{1}}{\mathrm{2}}\leqslant\:\mathrm{1}\leqslant\:{m}+\mathrm{1}\:\Rightarrow \\ $$$$−{m}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{m}\geqslant\mathrm{0}\:\Rightarrow{m}\geqslant\mathrm{0}\Rightarrow\:{m}\geqslant\mathrm{1} \\ $$$${if}\:−\mathrm{1}<{m}<\mathrm{1}\:\Rightarrow−\left(\mathrm{1}−{m}\right)\leqslant\sqrt{{m}^{\mathrm{2}} +\mathrm{2}}\leqslant{m}+\mathrm{1}\:\Rightarrow \\ $$$${m}^{\mathrm{2}} −\mathrm{2}{m}\:+\mathrm{1}\:\leqslant{m}^{\mathrm{2}} +\mathrm{2}\:\leqslant{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}\:\Rightarrow \\ $$$$−{m}\leqslant\frac{\mathrm{1}}{\mathrm{2}}{and}\:{m}\geqslant\mathrm{0}\:\Rightarrow{m}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{0}\leqslant{m}\leqslant\mathrm{1} \\ $$$$\mid{t}_{\mathrm{2}} \mid\leqslant\mathrm{1}\:\Rightarrow\:\mid−{m}−\sqrt{{m}^{\mathrm{2}} \:+\mathrm{2}}\mid\leqslant\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{1}\leqslant{m}+\sqrt{{m}^{\mathrm{2}} \:+\mathrm{2}}\leqslant\mathrm{1}\:\Rightarrow−{m}−\mathrm{1}\leqslant\sqrt{{m}^{\mathrm{2}} \:+\mathrm{2}}\leqslant\mathrm{1}−{m}\:{so} \\ $$$${m}\leqslant\mathrm{1}\:\:{if}\:−{m}−\mathrm{1}>\mathrm{0}\:\Rightarrow{m}\leqslant−\mathrm{1}\:\Rightarrow \\ $$$${m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}\leqslant{m}^{\mathrm{2}} +\mathrm{2}\leqslant{m}^{\mathrm{2}} −\mathrm{2}{m}\:+\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}{m}+\mathrm{1}\leqslant\mathrm{2}\leqslant−\mathrm{2}{m}+\mathrm{1}\:\Rightarrow\:\mathrm{2}{m}\leqslant\mathrm{1}\leqslant−\mathrm{2}{m}\:\Rightarrow \\ $$$${m}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant−{m}\:\Rightarrow{m}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{m}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${m}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{m}\:\leqslant−\mathrm{1}\:{due}\:{to}\:{initial}\:{condition} \\ $$$${if}\:−{m}−\mathrm{1}\leqslant\mathrm{0}\:\Rightarrow−\mathrm{1}\leqslant{m}\leqslant\mathrm{1}\:\:{so} \\ $$$$\mid{t}_{\mathrm{2}} \mid\:\leqslant\mathrm{1}\:\Rightarrow\:−\mathrm{1}\leqslant{m}\leqslant\mathrm{1}\:{finally} \\ $$$$\mid{t}_{\mathrm{1}} \mid\leqslant\mathrm{1}\:{and}\:\mid{t}_{\mathrm{2}} \mid\leqslant\mathrm{1}\:\Rightarrow\:\:\:\mathrm{0}\leqslant{m}\leqslant\mathrm{1}\:. \\ $$
Commented by math khazana by abdo last updated on 15/Aug/18
another way but easy we have ∣t_1 ∣≤1 and ∣t_2 ∣≤1 ⇒ −1 ≤−m+(√(m^2 +2))≤1 and −1≤−m−(√(m^2 +2))≤1 ⇒ −2≤−2m ≤2 ⇒ −1≤−m≤1 ⇒ −1≤m≤1
$${another}\:{way}\:{but}\:{easy}\:\:{we}\:{have}\:\mid{t}_{\mathrm{1}} \mid\leqslant\mathrm{1}\:{and} \\ $$$$\mid{t}_{\mathrm{2}} \mid\leqslant\mathrm{1}\:\Rightarrow\:−\mathrm{1}\:\leqslant−{m}+\sqrt{{m}^{\mathrm{2}} \:+\mathrm{2}}\leqslant\mathrm{1}\:{and} \\ $$$$−\mathrm{1}\leqslant−{m}−\sqrt{{m}^{\mathrm{2}} \:+\mathrm{2}}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{2}\leqslant−\mathrm{2}{m}\:\leqslant\mathrm{2}\:\Rightarrow\:−\mathrm{1}\leqslant−{m}\leqslant\mathrm{1}\:\Rightarrow\:−\mathrm{1}\leqslant{m}\leqslant\mathrm{1} \\ $$
Commented by math khazana by abdo last updated on 15/Aug/18
−1≤m≤1
$$−\mathrm{1}\leqslant{m}\leqslant\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 15/Aug/18
for m=1 h(x)=(√(sin^4 x+cos^4 x−2sinxcosx)) =(√((sin^2 x +cos^2 x)^2 −2sin^2 xcos^2 x−sin(2x))) =(√(1−(1/2)sin^2 (2x)−sin(2x)))=(1/( (√2)))(√(2−sin^2 (2x)−2sin(2x))) =(1/( (√2)))(√(−sin^2 (2x)−2sin(2x)+2)) let t=sin(2x) Δ^′ =1 +2 =3 ⇒ t_1 =((1+(√3))/(−1)) and t_2 =((1−(√3))/(−1)) ⇒t_1 =−(√3)−1 and t_2 =−1−(√3) so ∣t_1 ∣<1 but ∣t_2 ∣ >1 any way we can take −(1/2)≤m≤(1/2) thank you sir for this remark...
$${for}\:{m}=\mathrm{1}\:\:{h}\left({x}\right)=\sqrt{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}−\mathrm{2}{sinxcosx}} \\ $$$$=\sqrt{\left({sin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}−{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−{sin}\left(\mathrm{2}{x}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\mathrm{2}−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)+\mathrm{2}}\:\:{let}\:{t}={sin}\left(\mathrm{2}{x}\right) \\ $$$$\Delta^{'} \:=\mathrm{1}\:+\mathrm{2}\:=\mathrm{3}\:\Rightarrow\:{t}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{−\mathrm{1}}\:\:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{−\mathrm{1}}\:\Rightarrow{t}_{\mathrm{1}} =−\sqrt{\mathrm{3}}−\mathrm{1}\:\:\:{and} \\ $$$${t}_{\mathrm{2}} =−\mathrm{1}−\sqrt{\mathrm{3}}\:\:\:\:{so}\:\mid{t}_{\mathrm{1}} \mid<\mathrm{1}\:\:{but}\:\mid{t}_{\mathrm{2}} \mid\:>\mathrm{1}\:\:{any}\:{way}\:{we}\:{can}\:{take}\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{m}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${thank}\:{you}\:{sir}\:{for}\:{this}\:{remark}… \\ $$
Answered by MJS last updated on 15/Aug/18
sin^4 x +cos^4 x −2msin x cos x ≥0 ...after some trigonometric transformations: (1/4)cos 4x −msin 2x +(3/4)≥0 (d/dx)[(1/4)cos 4x −msin 2x +(3/4)]=0 −sin 4x −2mcos 2x=0 x=arctan t ((2mt^4 +4t^3 −4t^2 −2m)/((t^2 +1)^2 ))=0 t^4 +((2t^3 )/m)−((2t)/m)−1=0 t=±1 ∨ t=−(1/m)±((√(1−m^2 ))/m) ⇒ −1≤m≤1 ⇒ x=±(π/4) ∨ x=arctan (−(1/m)±((√(1−m^2 ))/m)) (1/4)cos 4x −msin 2x +(3/4)≥0 x_1 =−(π/4) ⇒ m+(1/2)≥0 ⇒ m≥−(1/2) x_2 =(π/4) ⇒ −m+(1/2)≥0 ⇒ m≤(1/2) x_(3, 4) =arctan (−(1/m)±((√(1−m^2 ))/m)) ⇒ (m^2 /2)+1≥0 ⇒ m∈R answer is −(1/2)≤m≤(1/2)
$$\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x}\:−\mathrm{2}{m}\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:\geqslant\mathrm{0} \\ $$$$…\mathrm{after}\:\mathrm{some}\:\mathrm{trigonometric}\:\mathrm{transformations}: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{4}{x}\:−{m}\mathrm{sin}\:\mathrm{2}{x}\:+\frac{\mathrm{3}}{\mathrm{4}}\geqslant\mathrm{0} \\ $$$$\frac{{d}}{{dx}}\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{4}{x}\:−{m}\mathrm{sin}\:\mathrm{2}{x}\:+\frac{\mathrm{3}}{\mathrm{4}}\right]=\mathrm{0} \\ $$$$−\mathrm{sin}\:\mathrm{4}{x}\:−\mathrm{2}{m}\mathrm{cos}\:\mathrm{2}{x}=\mathrm{0} \\ $$$${x}=\mathrm{arctan}\:{t} \\ $$$$\frac{\mathrm{2}{mt}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{3}} −\mathrm{4}{t}^{\mathrm{2}} −\mathrm{2}{m}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${t}^{\mathrm{4}} +\frac{\mathrm{2}{t}^{\mathrm{3}} }{{m}}−\frac{\mathrm{2}{t}}{{m}}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\pm\mathrm{1}\:\vee\:{t}=−\frac{\mathrm{1}}{{m}}\pm\frac{\sqrt{\mathrm{1}−{m}^{\mathrm{2}} }}{{m}}\:\Rightarrow\:−\mathrm{1}\leqslant{m}\leqslant\mathrm{1} \\ $$$$\Rightarrow\:{x}=\pm\frac{\pi}{\mathrm{4}}\:\vee\:{x}=\mathrm{arctan}\:\left(−\frac{\mathrm{1}}{{m}}\pm\frac{\sqrt{\mathrm{1}−{m}^{\mathrm{2}} }}{{m}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{4}{x}\:−{m}\mathrm{sin}\:\mathrm{2}{x}\:+\frac{\mathrm{3}}{\mathrm{4}}\geqslant\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\frac{\pi}{\mathrm{4}}\:\Rightarrow\:{m}+\frac{\mathrm{1}}{\mathrm{2}}\geqslant\mathrm{0}\:\Rightarrow\:{m}\geqslant−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\frac{\pi}{\mathrm{4}}\:\Rightarrow\:−{m}+\frac{\mathrm{1}}{\mathrm{2}}\geqslant\mathrm{0}\:\Rightarrow\:{m}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =\mathrm{arctan}\:\left(−\frac{\mathrm{1}}{{m}}\pm\frac{\sqrt{\mathrm{1}−{m}^{\mathrm{2}} }}{{m}}\right)\:\Rightarrow\:\frac{{m}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow\:{m}\in\mathbb{R} \\ $$$$\mathrm{answer}\:\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{m}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by lucha116 last updated on 15/Aug/18
it is true but i have studied grade 11 and i have not learnt (d/dx). Can you solve by other way?
$${it}\:{is}\:{true}\:{but}\:{i}\:{have}\:{studied}\:{grade}\:\mathrm{11}\:{and}\:{i}\:{have}\:{not}\:{learnt}\:\frac{{d}}{{dx}}.\:{Can}\:{you}\:{solve}\:{by}\:{other}\:{way}? \\ $$
Commented by maxmathsup by imad last updated on 15/Aug/18
this question is more difficult for grade 11 and generally the consept of trigonometry is more difficult ...
$${this}\:{question}\:{is}\:{more}\:{difficult}\:{for}\:{grade}\:\mathrm{11}\:{and}\:{generally}\:{the}\:{consept}\:{of}\: \\ $$$${trigonometry}\:{is}\:{more}\:{difficult}\:\:… \\ $$
Commented by lucha116 last updated on 15/Aug/18
yes. i know. thks ^� 0 ^�
$${yes}.\:{i}\:{know}.\:{thks}\hat {\:}\mathrm{0}\hat {\:} \\ $$
Answered by MJS last updated on 15/Aug/18
hopefully an easier explanation sin^2 x =(1/2)(1−cos 2x) cos^2 x =(1/2)(1+cos 2x) sin^4 x =((1/2)(1−cos 2x))^2 = =(1/4)(1−2cos 2x +cos^2 2x)= =(1/4)(1−2cos 2x +(1/2)(1+cos 4x))= =(3/8)+(1/8)cos 4x −(1/2)cos 2x similar cos^4 x =(3/8)+(1/8)cos 4x +(1/2)cos 2x ⇒ sin^4 x +cos^4 x =(3/4)+(1/4)cos 4x this has a period of (π/2) with min= ((((π/4)+(π/2)z)),((1/2)) ) ; z∈Z −2msin x cos x =−msin 2x this has a period of π with min= ((( (π/4)+πz)),((−m)) ) ; z∈Z the sum of both gives (((π/4)),(((1/2)−m)) ) within the first period. we know that (1/2)−m≥0 ⇒ m≤(1/2) if m<0 the min of −msin 2x changes to the 2^(nd) half of its period because the curve now is upside down: min= (((((3π)/4)+πz)),(m) ) so the sum now is ((((3π)/4)),(((1/2)+m)) ) ⇒ m≥−(1/2)
$$\mathrm{hopefully}\:\mathrm{an}\:\mathrm{easier}\:\mathrm{explanation} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right) \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{x}\:=\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)\right)^{\mathrm{2}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{2cos}\:\mathrm{2}{x}\:+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{2cos}\:\mathrm{2}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{4}{x}\right)\right)= \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\mathrm{similar}\:\mathrm{cos}^{\mathrm{4}} \:{x}\:=\frac{\mathrm{3}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\Rightarrow\:\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x}\:=\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{4}{x} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:\frac{\pi}{\mathrm{2}}\:\mathrm{with}\:\mathrm{min}=\begin{pmatrix}{\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}{z}}\\{\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix}\:;\:{z}\in\mathbb{Z} \\ $$$$−\mathrm{2}{m}\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:=−{m}\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:\pi\:\mathrm{with}\:\mathrm{min}=\begin{pmatrix}{\:\frac{\pi}{\mathrm{4}}+\pi{z}}\\{−{m}}\end{pmatrix}\:;\:{z}\in\mathbb{Z} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{both}\:\mathrm{gives}\:\begin{pmatrix}{\frac{\pi}{\mathrm{4}}}\\{\frac{\mathrm{1}}{\mathrm{2}}−{m}}\end{pmatrix}\:\mathrm{within}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{period}.\:\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\frac{\mathrm{1}}{\mathrm{2}}−{m}\geqslant\mathrm{0}\:\Rightarrow\:{m}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{if}\:{m}<\mathrm{0}\:\mathrm{the}\:\mathrm{min}\:\mathrm{of}\:−{m}\mathrm{sin}\:\mathrm{2}{x}\:\mathrm{changes}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{half}\:\mathrm{of}\:\mathrm{its}\:\mathrm{period}\:\mathrm{because}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{now} \\ $$$$\mathrm{is}\:\mathrm{upside}\:\mathrm{down}:\:\mathrm{min}=\begin{pmatrix}{\frac{\mathrm{3}\pi}{\mathrm{4}}+\pi{z}}\\{{m}}\end{pmatrix} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{now}\:\mathrm{is}\:\begin{pmatrix}{\frac{\mathrm{3}\pi}{\mathrm{4}}}\\{\frac{\mathrm{1}}{\mathrm{2}}+{m}}\end{pmatrix}\:\:\Rightarrow\:{m}\geqslant−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *