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


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Question Number 214658 by ChantalYah last updated on 15/Dec/24

$1) The function H is defined by H (x) = 3 \cosh \frac{x}{3} + \sinh \frac{x}{3} .$ $Find the value of λ for which H (\ln λ^{3}) = 4$ $2) Prove that \int_{2}^{4} \frac{6 x + 1}{(2 x - 3) (3 x - 2)} dx = \ln 10 .$ $3) Show that \frac{\sin θ + \sin 2 θ}{1 + \cos θ + \cos 2 θ} \equiv \tan θ$ $4) If z = \cos θ + i \sin θ, Show that z + \frac{1}{z} = 2 \cos θ and that$ $z^{n} + \frac{1}{z^{n}} = 2 \cos n θ hence or otherwise show that$ ${32 \cos}^{6} θ = \cos 6 θ + 6 \cos 4 θ + 15 \cos 2 θ + 10 .$ $Mr Hans$
	Answered by a.lgnaoui last updated on 16/Dec/24
	$2)∫_2 ^4 ((6x+1)/((2x−3)(3x−2)))dx= ((6x+1)/((2x−3)(3x−2)))=(a/(2x−3))+(b/((3x−2))) =(((3a+2b)x−(2a+3b))/((2x−3)(3x−2))) { ((3a+2b=6 ×2)),((2a+3b=−1 ×−3)) :} −5b=15 ⇒ a=4 b=−3 ∫_2 ^4 ((6x+1)/((2x−3)(3x+2)))= ∫(4/(2x−3))dx−∫(3/(3x−2))dx •2[ln(2x−3)]_2 ^4 −[ln(3x−2) −(ln(3x−2)]_2 ^4 = 2[ln(5)−(ln10)+ln4=ln5+ln2=ln10 •3)((sin 𝛉+sin 2𝛉)/(1+cos 𝛉+cos 2𝛉))=((sin𝛉(1+2cos 𝛉) )/(cos 𝛉(1+2cos 𝛉))) tan 𝛉 •4)if z=cos 𝛉+isin 𝛉 z+(1/z)=cos 𝛉+isin 𝛉+((cos 𝛉−isin 𝛉)/1) =2cos 𝛉 ⇒ z^n +(1/z^n )=2cos n𝛉 32cos 6𝛉=2^5 cos 6𝛉=2^4 ×(2cos 6𝛉) 2^4 [(cos 6𝛉+isin 6𝛉)+(1/(cos 6𝛉+isin 6𝛉)) =2^5 (2cos^2 3𝛉−1) cos 3𝛉=4cos^3 𝛉−3cos 𝛉 cos 6𝛉=2cos^2 3𝛉−1 =2cos^2 𝛉(4cos^2 𝛉−3)^2 −1 =2(16cos^6 𝛉−24cos^4 𝛉+9cos^2 𝛉)−1 soit 32cos^6 𝛉=cos 6𝛉+48cos^4 𝛉−18cos^2 𝛉+1 cos 4𝛉=2(2cos^2 𝛉−1)^2 −1 =2(4cos^4 𝛉+1−4cos^2 𝛉)−1 =8cos^4 𝛉−8cos^2 𝛉+1 ⇒ 48cos^4 𝛉−18cos^2 𝛉 =6cos 4𝛉+30cos^2 𝛉 32cos^6 𝛉=cos 6𝛉+6cos 4𝛉+30cos^2 𝛉 =cos 6𝛉+6cos 4𝛉+15(cos 2𝛉)+10 i$
	$2) \int_{2}^{4} \frac{6 x + 1}{(2 x - 3) (3 x - 2)} dx =$ $\frac{6 x + 1}{(2 x - 3) (3 x - 2)} = \frac{a}{2 x - 3} + \frac{b}{(3 x - 2)}$ $= \frac{(3 a + 2 b) x - (2 a + 3 b)}{(2 x - 3) (3 x - 2)}$ ${\begin{cases} 3 a + 2 b = 6 \times 2 \\ 2 a + 3 b = - 1 \times - 3 \end{cases}$ $- 5 b = 15 \Rightarrow a = 4 b = - 3$ $\int_{2}^{4} \frac{6 x + 1}{(2 x - 3) (3 x + 2)} =$ $\int \frac{4}{2 x - 3} dx - \int \frac{3}{3 x - 2} dx$ $∙ 2 {[\ln (2 x - 3)]}_{2}^{4} - [\ln (3 x - 2)$ $- {(\ln (3 x - 2)]}_{2}^{4} =$ $2 [\ln (5) - (\ln 10) + \ln 4 = \ln 5 + \ln 2 = \ln 10$ $∙ 3) \frac{\sin θ + \sin 2 θ}{1 + \cos θ + \cos 2 θ} = \frac{\sin θ (1 + 2 \cos θ)}{\cos θ (1 + 2 \cos θ)}$ $\tan θ$ $∙ 4) if z = \cos θ + isin θ$ $z + \frac{1}{z} = \cos θ + isin θ + \frac{\cos θ - isin θ}{1}$ $= 2 \cos θ \Rightarrow z^{n} + \frac{1}{z^{n}} = 2 \cos n θ$ $32 \cos 6 θ = 2^{5} \cos 6 θ = 2^{4} \times (2 \cos 6 θ)$ $2^{4} [(\cos 6 θ + isin 6 θ) + \frac{1}{\cos 6 θ + isin 6 θ}$ $= 2^{5} (2 \cos^{2} 3 θ - 1)$ $\cos 3 θ = 4 \cos^{3} θ - 3 \cos θ$ $\cos 6 θ = 2 \cos^{2} 3 θ - 1$ $= 2 \cos^{2} θ {(4 \cos^{2} θ - 3)}^{2} - 1$ $= 2 (16 \cos^{6} θ - 24 \cos^{4} θ + 9 \cos^{2} θ) - 1$ $soit$ $32 \cos^{6} θ = \cos 6 θ + 48 \cos^{4} θ - 18 \cos^{2} θ + 1$ $\cos 4 θ = 2 {(2 \cos^{2} θ - 1)}^{2} - 1$ $= 2 (4 \cos^{4} θ + 1 - 4 \cos^{2} θ) - 1$ $= 8 \cos^{4} θ - 8 \cos^{2} θ + 1$ $\Rightarrow 48 \cos^{4} θ - 18 \cos^{2} θ$ $= 6 \cos 4 θ + 30 \cos^{2} θ$ $32 \cos^{6} θ = \cos 6 θ + 6 \cos 4 θ + 30 \cos^{2} θ$ $= \cos 6 θ + 6 \cos 4 θ + 15 (\cos 2 θ) + 10$ $i$
	Answered by MathematicalUser2357 last updated on 16/Dec/24

	$\ln λ^{3} = 3 \ln λ$ $using hyperbolic trigonometric function idenities$ $H (x) = 3 \times \frac{e^{x / 3} - e^{- x / 3}}{2} + \frac{e^{x / 3} + e^{- x / 3}}{2}$ $H (x) = \frac{(3 e^{x / 3} - 3 e^{- x / 3}) + (e^{x / 3} + e^{- x / 3})}{2} = \frac{4 e^{x / 3} - 2 e^{- x / 3}}{2}$ $\begin{matrix} H (x) = 2 e^{x / 3} - e^{- x / 3}, H (3 \ln λ) = 2 λ - \frac{1}{λ} \end{matrix}$ $So 2 λ - \frac{1}{λ} = 4$ $2 λ^{2} - 4 λ - 1 = 0$ $λ = \frac{4 \pm \sqrt{4^{2} + 4 \times 2 \times 1}}{2 \times 2}$ $λ = \frac{4 \pm \sqrt{24}}{4} = \frac{4 \pm 2 \sqrt{6}}{4}$ $\begin{matrix} λ = 2 \pm \sqrt{6} \end{matrix} Question 1 is easy though$
	Commented by MathematicalUser2357 last updated on 17/Dec/24

	$Even though {it}^{'} s not a solution .$
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