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Using double angle and pythagorean identities (2 parts)

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Find the largest solution to each of the following equations, giving your answers in terms of natural logarithms.

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a)

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We'd like the arguments of the cosh functions to be the same, so we use the identity

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$\\cosh(2\\theta)=2\\cosh^2(\\theta)-1$

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To attain

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$(2\\cosh^2(\\var{a}x)-1)-\\cosh(\\var{a}x)=\\var{c}$

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$2\\cosh^2(\\var{a}x)\\cosh(\\var{a}x)-\\var{c+1}=0$.

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Proceed to solve this quadratic in $\\cosh(\\var{a}x)$ by factorisation or using the quadratic formula to attain

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$\\cosh(\\var{a}x)=\\var{(k+1)/2}, \\var{-k/2} .$

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Taking the only valid solution $\\var{(k+1)/2}$ (since $\\cosh(x)\\ge1$) and using the definition of $\\text{arcosh}(\\var{a}x)$, we now have

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$\\var{a}x=\\ln(\\var{(k+1)/2}+\\sqrt{(\\var{(k+1)/2})^2-1})$

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$x=\\frac{1}{\\var{a}}\\ln(\\var{(k+1)/2}+\\sqrt{\\var{((k+1)/2)^2-1}})$

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b)

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We begin by making use of the identity

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$\\cosh^2(x) - \\sinh^2(x) = 1$

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That is

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$\\sinh^2(x) = \\cosh^2(x) - 1$

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We can substitute this into the main equation

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$\\begin{aligned}
\\var{D}\\sinh^2(x) - \\var{B}\\cosh(x) &= \\var{B} \\\\
\\var{D}(\\cosh^2(x) - 1) - \\var{B}\\cosh(x) - \\var{B} &= 0 \\\\
\\var{D}\\cosh^2(x) - \\var{B}\\cosh(x) - \\var{B + D} &= 0
\\end{aligned}$

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This equation takes the form of a quadratic, so we subsitute $y = \\cosh(x)$ to get

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$\\var{D}y^2 - \\var{B}y - \\var{B+D} = 0$

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Then, by the quadratic formula (using $+$ to find the largest solution) we have

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$\\begin{aligned}
y &= \\frac{\\var{B} + \\sqrt{\\var{B^2} + \\simplify{ -4*{D}*(-{B}-{D})}}}{\\var{2D}} \\\\
&= \\frac{\\var{B} + \\sqrt{\\simplify{{B^2} + -4*{D}*(-{B}-{D})}}}{\\var{2D}} \\\\
&= \\frac{\\var{B} + \\simplify{{({B^2} + -4*{D}*(-{B}-{D}))^(1/2)}}}{\\var{2D}} \\\\
&= \\simplify{({B} + {({B^2} + -4*{D}*(-{B}-{D}))^(1/2)})/({2*D})} \\\\
\\end{aligned}$

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Therefore, finally we have

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$y = \\cosh(x) = \\frac{\\var{B + D}}{\\var{D}}$

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Which gives us

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$x = \\cosh^{-1}(y) = \\ln\\left( \\frac{\\var{B + D}}{\\var{D}} + \\sqrt{\\left(\\frac{\\var{B + D}}{\\var{D}}\\right)^2 - 1} \\right)$

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$\\cosh(\\var{2*a}x) - \\cosh(\\var{a}x) = \\var{c}$

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$x = $ [[0]]

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$\\var{D}\\sinh^2(x) - \\var{B}\\cosh(x) = \\var{B}$

\n

$x = $ [[0]]

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