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Question Number 214658 by ChantalYah last updated on 15/Dec/24
1) The function H is defined by H(x) =3cosh(x/3)+sinh(x/3). Find the value of λ for which H(lnλ^3 )=4 2) Prove that ∫_2 ^4 ((6x+1)/((2x−3)(3x−2)))dx= ln 10. 3)Show that ((sinθ + sin2θ)/(1+ cosθ+ cos2θ))≡ tanθ 4) If z=cosθ+ i sinθ, Show that z+(1/z)=2cosθ and that z^n +(1/z^n )=2cos nθ hence or otherwise show that 32cos^6 θ= cos 6θ +6cos 4θ + 15cos 2θ+ 10. Mr Hans
$$\left.\mathrm{1}\right)\:\mathrm{The}\:\mathrm{function}\:\mathrm{H}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{H}\left(\mathrm{x}\right)\:=\mathrm{3cosh}\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{sinh}\frac{\mathrm{x}}{\mathrm{3}}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\lambda\:\mathrm{for}\:\mathrm{which}\:\mathrm{H}\left(\mathrm{ln}\lambda^{\mathrm{3}} \right)=\mathrm{4} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\mathrm{6x}+\mathrm{1}}{\left(\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3x}−\mathrm{2}\right)}\mathrm{dx}=\:\mathrm{ln}\:\mathrm{10}. \\ $$$$\left.\mathrm{3}\right)\mathrm{Show}\:\mathrm{that}\:\frac{\mathrm{sin}\theta\:+\:\mathrm{sin2}\theta}{\mathrm{1}+\:\mathrm{cos}\theta+\:\mathrm{cos2}\theta}\equiv\:\mathrm{tan}\theta \\ $$$$\left.\mathrm{4}\right)\:\mathrm{If}\:\mathrm{z}=\mathrm{cos}\theta+\:\mathrm{i}\:\mathrm{sin}\theta,\:\mathrm{Show}\:\mathrm{that}\:\mathrm{z}+\frac{\mathrm{1}}{\mathrm{z}}=\mathrm{2cos}\theta\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{z}^{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{n}} }=\mathrm{2cos}\:\mathrm{n}\theta\:\mathrm{hence}\:\mathrm{or}\:\mathrm{otherwise}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{32cos}^{\mathrm{6}} \theta=\:\mathrm{cos}\:\mathrm{6}\theta\:+\mathrm{6cos}\:\mathrm{4}\theta\:+\:\mathrm{15cos}\:\mathrm{2}\theta+\:\mathrm{10}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Mr}\:\mathrm{Hans} \\ $$
Answered by a.lgnaoui last updated on 16/Dec/24
2)∫_2 ^4 ((6x+1)/((2x−3)(3x−2)))dx= ((6x+1)/((2x−3)(3x−2)))=(a/(2x−3))+(b/((3x−2))) =(((3a+2b)x−(2a+3b))/((2x−3)(3x−2))) { ((3a+2b=6 ×2)),((2a+3b=−1 ×−3)) :} −5b=15 ⇒ a=4 b=−3 ∫_2 ^4 ((6x+1)/((2x−3)(3x+2)))= ∫(4/(2x−3))dx−∫(3/(3x−2))dx •2[ln(2x−3)]_2 ^4 −[ln(3x−2) −(ln(3x−2)]_2 ^4 = 2[ln(5)−(ln10)+ln4=ln5+ln2=ln10 •3)((sin 𝛉+sin 2𝛉)/(1+cos 𝛉+cos 2𝛉))=((sin𝛉(1+2cos 𝛉) )/(cos 𝛉(1+2cos 𝛉))) tan 𝛉 •4)if z=cos 𝛉+isin 𝛉 z+(1/z)=cos 𝛉+isin 𝛉+((cos 𝛉−isin 𝛉)/1) =2cos 𝛉 ⇒ z^n +(1/z^n )=2cos n𝛉 32cos 6𝛉=2^5 cos 6𝛉=2^4 ×(2cos 6𝛉) 2^4 [(cos 6𝛉+isin 6𝛉)+(1/(cos 6𝛉+isin 6𝛉)) =2^5 (2cos^2 3𝛉−1) cos 3𝛉=4cos^3 𝛉−3cos 𝛉 cos 6𝛉=2cos^2 3𝛉−1 =2cos^2 𝛉(4cos^2 𝛉−3)^2 −1 =2(16cos^6 𝛉−24cos^4 𝛉+9cos^2 𝛉)−1 soit 32cos^6 𝛉=cos 6𝛉+48cos^4 𝛉−18cos^2 𝛉+1 cos 4𝛉=2(2cos^2 𝛉−1)^2 −1 =2(4cos^4 𝛉+1−4cos^2 𝛉)−1 =8cos^4 𝛉−8cos^2 𝛉+1 ⇒ 48cos^4 𝛉−18cos^2 𝛉 =6cos 4𝛉+30cos^2 𝛉 32cos^6 𝛉=cos 6𝛉+6cos 4𝛉+30cos^2 𝛉 =cos 6𝛉+6cos 4𝛉+15(cos 2𝛉)+10 i
$$\left.\mathrm{2}\right)\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\mathrm{6x}+\mathrm{1}}{\left(\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3x}−\mathrm{2}\right)}\mathrm{dx}= \\ $$$$\:\frac{\mathrm{6x}+\mathrm{1}}{\left(\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3x}−\mathrm{2}\right)}=\frac{\mathrm{a}}{\mathrm{2x}−\mathrm{3}}+\frac{\mathrm{b}}{\left(\mathrm{3x}−\mathrm{2}\right)} \\ $$$$=\frac{\left(\mathrm{3a}+\mathrm{2b}\right)\mathrm{x}−\left(\mathrm{2a}+\mathrm{3b}\right)}{\left(\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3x}−\mathrm{2}\right)} \\ $$$$\:\begin{cases}{\mathrm{3a}+\mathrm{2b}=\mathrm{6}\:\:\:\:×\mathrm{2}}\\{\mathrm{2a}+\mathrm{3b}=−\mathrm{1}\:\:×−\mathrm{3}}\end{cases} \\ $$$$\:\:−\mathrm{5b}=\mathrm{15}\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\mathrm{a}=\mathrm{4}\:\:\:\mathrm{b}=−\mathrm{3}\:\:\: \\ $$$$\:\:\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\mathrm{6x}+\mathrm{1}}{\left(\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3x}+\mathrm{2}\right)}= \\ $$$$\int\frac{\mathrm{4}}{\mathrm{2x}−\mathrm{3}}\mathrm{dx}−\int\frac{\mathrm{3}}{\mathrm{3x}−\mathrm{2}}\mathrm{dx} \\ $$$$ \\ $$$$\bullet\mathrm{2}\left[\mathrm{ln}\left(\mathrm{2x}−\mathrm{3}\right)\right]_{\mathrm{2}} ^{\mathrm{4}} −\left[\mathrm{ln}\left(\mathrm{3x}−\mathrm{2}\right)\right. \\ $$$$\:\:−\left(\mathrm{ln}\left(\mathrm{3x}−\mathrm{2}\right)\right]_{\mathrm{2}} ^{\mathrm{4}} = \\ $$$$\mathrm{2}\left[\mathrm{ln}\left(\mathrm{5}\right)−\left(\mathrm{ln10}\right)+\mathrm{ln4}=\mathrm{ln5}+\mathrm{ln2}=\mathrm{ln10}\right. \\ $$$$ \\ $$$$\left.\bullet\mathrm{3}\right)\frac{\mathrm{sin}\:\boldsymbol{\theta}+\mathrm{sin}\:\mathrm{2}\boldsymbol{\theta}}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}+\mathrm{cos}\:\mathrm{2}\boldsymbol{\theta}}=\frac{\mathrm{sin}\boldsymbol{\theta}\left(\mathrm{1}+\mathrm{2cos}\:\boldsymbol{\theta}\right)\:}{\mathrm{cos}\:\boldsymbol{\theta}\left(\mathrm{1}+\mathrm{2cos}\:\boldsymbol{\theta}\right)} \\ $$$$\:\:\:\mathrm{tan}\:\boldsymbol{\theta} \\ $$$$\:\: \\ $$$$\left.\bullet\mathrm{4}\right)\mathrm{if}\:\:\:\boldsymbol{\mathrm{z}}=\mathrm{cos}\:\boldsymbol{\theta}+\mathrm{isin}\:\boldsymbol{\theta} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{z}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}=\mathrm{cos}\:\boldsymbol{\theta}+\mathrm{isin}\:\boldsymbol{\theta}+\frac{\mathrm{cos}\:\boldsymbol{\theta}−\mathrm{isin}\:\boldsymbol{\theta}}{\mathrm{1}} \\ $$$$\:\:\:=\mathrm{2cos}\:\boldsymbol{\theta}\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{z}}^{\mathrm{n}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}^{\boldsymbol{\mathrm{n}}} }=\mathrm{2cos}\:\boldsymbol{\mathrm{n}\theta} \\ $$$$\:\mathrm{32cos}\:\mathrm{6}\boldsymbol{\theta}=\mathrm{2}^{\mathrm{5}} \mathrm{cos}\:\mathrm{6}\boldsymbol{\theta}=\mathrm{2}^{\mathrm{4}} ×\left(\mathrm{2cos}\:\mathrm{6}\boldsymbol{\theta}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{4}} \left[\left(\mathrm{cos}\:\mathrm{6}\boldsymbol{\theta}+\mathrm{isin}\:\mathrm{6}\boldsymbol{\theta}\right)+\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}\boldsymbol{\theta}+\mathrm{isin}\:\mathrm{6}\boldsymbol{\theta}}\right. \\ $$$$\:\:\:=\mathrm{2}^{\mathrm{5}} \left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{3}\boldsymbol{\theta}−\mathrm{1}\right) \\ $$$$\mathrm{cos}\:\mathrm{3}\boldsymbol{\theta}=\mathrm{4cos}\:^{\mathrm{3}} \boldsymbol{\theta}−\mathrm{3cos}\:\boldsymbol{\theta} \\ $$$$\:\mathrm{cos}\:\mathrm{6}\boldsymbol{\theta}=\mathrm{2cos}\:^{\mathrm{2}} \mathrm{3}\boldsymbol{\theta}−\mathrm{1} \\ $$$$=\mathrm{2cos}\:^{\mathrm{2}} \boldsymbol{\theta}\left(\mathrm{4cos}\:^{\mathrm{2}} \boldsymbol{\theta}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\:=\mathrm{2}\left(\mathrm{16cos}\:^{\mathrm{6}} \boldsymbol{\theta}−\mathrm{24cos}\:^{\mathrm{4}} \boldsymbol{\theta}+\mathrm{9cos}\:^{\mathrm{2}} \boldsymbol{\theta}\right)−\mathrm{1} \\ $$$$\mathrm{soit} \\ $$$$\:\:\:\mathrm{32cos}\:^{\mathrm{6}} \boldsymbol{\theta}=\mathrm{cos}\:\mathrm{6}\boldsymbol{\theta}+\mathrm{48cos}\:^{\mathrm{4}} \boldsymbol{\theta}−\mathrm{18cos}\:^{\mathrm{2}} \boldsymbol{\theta}+\mathrm{1} \\ $$$$ \\ $$$$\mathrm{cos}\:\mathrm{4}\boldsymbol{\theta}=\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \boldsymbol{\theta}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$=\mathrm{2}\left(\mathrm{4cos}\:^{\mathrm{4}} \boldsymbol{\theta}+\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \boldsymbol{\theta}\right)−\mathrm{1} \\ $$$$=\mathrm{8cos}\:^{\mathrm{4}} \boldsymbol{\theta}−\mathrm{8cos}\:^{\mathrm{2}} \boldsymbol{\theta}+\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:\:\mathrm{48cos}\:^{\mathrm{4}} \boldsymbol{\theta}−\mathrm{18cos}\:^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$=\mathrm{6cos}\:\mathrm{4}\boldsymbol{\theta}+\mathrm{30cos}\:^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$\mathrm{32cos}\:^{\mathrm{6}} \boldsymbol{\theta}=\mathrm{cos}\:\mathrm{6}\boldsymbol{\theta}+\mathrm{6cos}\:\mathrm{4}\boldsymbol{\theta}+\mathrm{30cos}\:^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$=\mathrm{cos}\:\mathrm{6}\boldsymbol{\theta}+\mathrm{6cos}\:\mathrm{4}\boldsymbol{\theta}+\mathrm{15}\left(\mathrm{cos}\:\mathrm{2}\boldsymbol{\theta}\right)+\mathrm{10} \\ $$$$\mathrm{i} \\ $$
Answered by MathematicalUser2357 last updated on 16/Dec/24
ln λ^3 =3ln λ using hyperbolic trigonometric function idenities H(x)=3×((e^(x/3) −e^(−x/3) )/2)+((e^(x/3) +e^(−x/3) )/2) H(x)=(((3e^(x/3) −3e^(−x/3) )+(e^(x/3) +e^(−x/3) ))/2)=((4e^(x/3) −2e^(−x/3) )/2) determinant (((H(x)=2e^(x/3) −e^(−x/3) , H(3ln λ)=2λ−(1/λ)))) So 2λ−(1/λ)=4 2λ^2 −4λ−1=0 λ=((4±(√(4^2 +4×2×1)))/(2×2)) λ=((4±(√(24)))/4)=((4±2(√6))/4) determinant (((λ=2±(√6))))Question 1 is easy though
$$\mathrm{ln}\:\lambda^{\mathrm{3}} =\mathrm{3ln}\:\lambda \\ $$$$\mathrm{using}\:\mathrm{hyperbolic}\:\mathrm{trigonometric}\:\mathrm{function}\:\mathrm{idenities} \\ $$$$\mathrm{H}\left({x}\right)=\mathrm{3}×\frac{{e}^{{x}/\mathrm{3}} −{e}^{−{x}/\mathrm{3}} }{\mathrm{2}}+\frac{{e}^{{x}/\mathrm{3}} +{e}^{−{x}/\mathrm{3}} }{\mathrm{2}} \\ $$$$\mathrm{H}\left({x}\right)=\frac{\left(\mathrm{3}{e}^{{x}/\mathrm{3}} −\mathrm{3}{e}^{−{x}/\mathrm{3}} \right)+\left({e}^{{x}/\mathrm{3}} +{e}^{−{x}/\mathrm{3}} \right)}{\mathrm{2}}=\frac{\mathrm{4}{e}^{{x}/\mathrm{3}} −\mathrm{2}{e}^{−{x}/\mathrm{3}} }{\mathrm{2}} \\ $$$$\begin{array}{|c|}{\mathrm{H}\left({x}\right)=\mathrm{2}{e}^{{x}/\mathrm{3}} −{e}^{−{x}/\mathrm{3}} ,\:\mathrm{H}\left(\mathrm{3ln}\:\lambda\right)=\mathrm{2}\lambda−\frac{\mathrm{1}}{\lambda}}\\\hline\end{array} \\ $$$$\mathrm{So}\:\mathrm{2}\lambda−\frac{\mathrm{1}}{\lambda}=\mathrm{4} \\ $$$$\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{4}\lambda−\mathrm{1}=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{4}\pm\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{4}×\mathrm{2}×\mathrm{1}}}{\mathrm{2}×\mathrm{2}} \\ $$$$\lambda=\frac{\mathrm{4}\pm\sqrt{\mathrm{24}}}{\mathrm{4}}=\frac{\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{4}} \\ $$$$\begin{array}{|c|}{\lambda=\mathrm{2}\pm\sqrt{\mathrm{6}}}\\\hline\end{array}\mathrm{Question}\:\mathrm{1}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{though} \\ $$
Commented by MathematicalUser2357 last updated on 17/Dec/24
Even though it′s not a solution.
$$\mathrm{Even}\:\mathrm{though}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{a}\:\mathrm{solution}. \\ $$

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