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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) \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 .

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