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Hyperbolic functions and identitiesEdexcel International A Level Further Maths: Flashcards

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$\sinh x$ in terms of exponentials?

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sinh⁡x\sinh x in terms of exponentials?
ex−e−x2\frac{e^x-e^{-x}}{2}
cosh⁡x\cosh x in terms of exponentials?
ex+e−x2\frac{e^x+e^{-x}}{2}
tanh⁡x\tanh x in terms of sinh⁡x\sinh x and cosh⁡x\cosh x, and in exponentials?
sinh⁡xcosh⁡x=ex−e−xex+e−x=e2x−1e2x+1\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}
Define sech⁡x\operatorname{sech}x, cosech⁡x\operatorname{cosech}x, coth⁡x\coth x.
1cosh⁡x\frac{1}{\cosh x}, 1sinh⁡x\frac{1}{\sinh x}, 1tanh⁡x\frac{1}{\tanh x}
Range of cosh⁡x\cosh x?
y≥1y\ge1, with minimum 11 at x=0x=0.
Range of tanh⁡x\tanh x and its asymptotes?
−1<y<1-1<y<1; asymptotes y=1y=1 and y=−1y=-1.
Which of sinh⁡\sinh, cosh⁡\cosh, tanh⁡\tanh are odd?
sinh⁡x\sinh x and tanh⁡x\tanh x are odd; cosh⁡x\cosh x is even.
Key identity linking cosh⁡2x\cosh^2x and sinh⁡2x\sinh^2x?
cosh⁡2x−sinh⁡2x=1\cosh^2x-\sinh^2x=1
cosh⁡2x\cosh2x in terms of cosh⁡2x\cosh^2x and sinh⁡2x\sinh^2x?
cosh⁡2x=cosh⁡2x+sinh⁡2x\cosh2x=\cosh^2x+\sinh^2x
cosh⁡2x\cosh2x in terms of sinh⁡x\sinh x only?
cosh⁡2x=1+2sinh⁡2x\cosh2x=1+2\sinh^2x
1−tanh⁡2x1-\tanh^2x?
sech⁡2x\operatorname{sech}^2x
First step for acosh⁡x+bsinh⁡x=ca\cosh x+b\sinh x=c?
Write in exponentials, multiply by exe^x and solve the quadratic in exe^x.
Why must you check roots of a quadratic in exe^x?
ex>0e^x>0, so only positive roots give a solution.
cosh⁡(ln⁡3)\cosh(\ln3)?
12(3+13)=53\frac12\left(3+\frac13\right)=\frac53

Exam questions on Hyperbolic functions and identities

  1. A student evaluates hyperbolic functions at x=ln⁡3x=\ln3, using sinh⁡x=ex−e−x2\sinh x=\frac{e^x-e^{-x}}{2} and cosh⁡x=ex+e−x2\cosh x=\frac{e^x+e^{-x}}{2}.
    Use the identity cosh⁡2x=cosh⁡2x+sinh⁡2x\cosh2x=\cosh^2x+\sinh^2x to find the exact value of cosh⁡(2ln⁡3)\cosh(2\ln3).2 marks
  2. The function ff is defined by f(x)=tanh⁡xf(x)=\tanh x for all real xx.
    Show that tanh⁡x=e2x−1e2x+1\tanh x=\frac{e^{2x}-1}{e^{2x}+1}.2 marks
  3. For real xx, sinh⁡x=ex−e−x2\sinh x=\frac{e^x-e^{-x}}{2} and cosh⁡x=ex+e−x2\cosh x=\frac{e^x+e^{-x}}{2}.
    Prove that cosh⁡2x−sinh⁡2x=1\cosh^2x-\sinh^2x=1.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).