Hyperbolic functions Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Hyperbolic functions questions. See exactly how to solve problems on hyperbolic-functions, exponential-definitions, identities, osborns-rule.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Find the exact value of sinh⁡(ln⁡(2))\sinh{\left(\ln{\left(2\right)}\right)}.

Worked solution

  1. Write down the expression to be evaluated

    sinh⁡(ln⁡(2))\sinh{\left(\ln{\left(2\right)}\right)}

    Start from the expression given in the question.

  2. Quote the exponential definition

    sinh⁡(x)=ex−e−x2\sinh(x)=\frac{e^{x}-e^{-x}}{2}

    Every hyperbolic function is defined directly in terms of exe^{x}.

  3. State the exact value

    sinh⁡(ln⁡(2))=34\sinh{\left(\ln{\left(2\right)}\right)}=\frac{3}{4}

    This is the exact value of the expression.

Answer
34\frac{3}{4}
Question 2
2 markseasy
Find the exact value of cosh⁡(ln⁡(3))\cosh{\left(\ln{\left(3\right)}\right)}.

Worked solution

  1. Write down the expression to be evaluated

    cosh⁡(ln⁡(3))\cosh{\left(\ln{\left(3\right)}\right)}

    Start from the expression given in the question.

  2. Quote the exponential definition

    cosh⁡(x)=ex+e−x2\cosh(x)=\frac{e^{x}+e^{-x}}{2}

    Every hyperbolic function is defined directly in terms of exe^{x}.

  3. Substitute the argument into the definition

    cosh⁡(ln⁡(3))=eln⁡(3)+e−ln⁡(3)2\cosh{\left(\ln{\left(3\right)}\right)}=\frac{e^{\ln{\left(3\right)}}+e^{-\ln{\left(3\right)}}}{2}

    Replace xx by the argument given in the question.

  4. State the exact value

    cosh⁡(ln⁡(3))=53\cosh{\left(\ln{\left(3\right)}\right)}=\frac{5}{3}

    This is the exact value of the expression.

Answer
53\frac{5}{3}
Question 3
2 markseasy
Find the exact value of tanh⁡(ln⁡(2))\tanh{\left(\ln{\left(2\right)}\right)}.

Worked solution

  1. Write down the expression to be evaluated

    tanh⁡(ln⁡(2))\tanh{\left(\ln{\left(2\right)}\right)}

    Start from the expression given in the question.

  2. Quote the exponential definition

    tanh⁡(x)=ex−e−xex+e−x\tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

    Every hyperbolic function is defined directly in terms of exe^{x}.

  3. State the exact value

    tanh⁡(ln⁡(2))=35\tanh{\left(\ln{\left(2\right)}\right)}=\frac{3}{5}

    This is the exact value of the expression.

Answer
35\frac{3}{5}
Question 4
2 markseasy
Find the exact value of cosh⁡(ln⁡(5))\cosh{\left(\ln{\left(5\right)}\right)}.

Worked solution

  1. Write down the expression to be evaluated

    cosh⁡(ln⁡(5))\cosh{\left(\ln{\left(5\right)}\right)}

    Start from the expression given in the question.

  2. Quote the exponential definition

    cosh⁡(x)=ex+e−x2\cosh(x)=\frac{e^{x}+e^{-x}}{2}

    Every hyperbolic function is defined directly in terms of exe^{x}.

  3. State the exact value

    cosh⁡(ln⁡(5))=135\cosh{\left(\ln{\left(5\right)}\right)}=\frac{13}{5}

    This is the exact value of the expression.

Answer
135\frac{13}{5}
Question 5
2 markseasy
Find the exact value of tanh⁡(ln⁡(3))\tanh{\left(\ln{\left(3\right)}\right)}.

Worked solution

  1. Write down the expression to be evaluated

    tanh⁡(ln⁡(3))\tanh{\left(\ln{\left(3\right)}\right)}

    Start from the expression given in the question.

  2. Quote the exponential definition

    tanh⁡(x)=ex−e−xex+e−x\tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

    Every hyperbolic function is defined directly in terms of exe^{x}.

  3. Substitute the argument into the definition

    tanh⁡(ln⁡(3))=eln⁡(3)−e−ln⁡(3)eln⁡(3)+e−ln⁡(3)\tanh{\left(\ln{\left(3\right)}\right)}=\frac{e^{\ln{\left(3\right)}}-e^{-\ln{\left(3\right)}}}{e^{\ln{\left(3\right)}}+e^{-\ln{\left(3\right)}}}

    Replace xx by the argument given in the question.

  4. State the exact value

    tanh⁡(ln⁡(3))=45\tanh{\left(\ln{\left(3\right)}\right)}=\frac{4}{5}

    This is the exact value of the expression.

Answer
45\frac{4}{5}

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