amazon.co.ukHyperbolic functions for Edexcel 9FM0 Core Pure 2 and OCR MEI: sinh, cosh and tanh, key identities and Osborn's rule, inverse functions as logarithms, differentiation, integration and catenary arc lengths - revision with worked examples and mark-scheme logic.
Definitions and graphs. For real x: sinh x = (e^x - e^-x)/2, cosh x = (e^x + e^-x)/2, tanh x = sinh x / cosh x. Read these as exponential averages: sinh is odd (sinh(-x) = -sinh x) and grows exponentially; cosh is even, always at least 1, with the catenary shape and minimum (0, 1); tanh is odd, lying between -1 and 1 with horizontal asymptotes y = ±1, so its range is (-1, 1). Adding and subtracting the definitions gives the two identities that unlock everything: e^x = cosh x + sinh x and e^-x = cosh x - sinh x - so knowing cosh x, sinh x and the sign of x determines x itself through a logarithm.
Core identities. From the definitions directly: cosh^2 x - sinh^2 x = 1. Dividing by cosh^2 x gives 1 - tanh^2 x = sech^2 x; similarly coth^2 x - 1 = cosech^2 x. Addition formulas (note the sign pattern: cosh behaves like trig cos, sinh like trig sin): sinh(A ± B) = sinh A cosh B ± cosh A sinh B and cosh(A ± B) = cosh A cosh B ± sinh A sinh B. Double angles: sinh 2x = 2 sinh x cosh x; cosh 2x = cosh^2 x + sinh^2 x = 2cosh^2 x - 1 = 1 + 2 sinh^2 x. The half-identities used constantly in integration: sinh^2 x = (cosh 2x - 1)/2 and cosh^2 x = (cosh 2x + 1)/2.
Osborn's rule. To convert any trigonometric identity into a hyperbolic one: replace cos by cosh and sin by sinh, then change the sign of every term containing a product of two sinh factors, explicit or hidden inside a squared term. Example: cos 2A = 1 - 2 sin^2 A becomes cosh 2x = 1 + 2 sinh^2 x (the sinh^2 term is an implied product of two sinh, so the sign flips); sin 2A = 2 sin A cos A becomes sinh 2x = 2 sinh x cosh x (only a sinh-cosh product, so no flip).
Worked example. Show that tanh x = (e^(2x) - 1)/(e^(2x) + 1) and state the range of tanh. Divide numerator and denominator of (e^x - e^-x)/(e^x + e^-x) by e^-x: (e^(2x) - 1)/(e^(2x) + 1). As x to infinity the ratio tends to 1; as x to -infinity it tends to -1; by oddness the range is (-1, 1). This form is the fastest route to solving tanh equations and deriving the artanh logarithm (Section 2).
When a question says using the definitions of sinh and cosh, you must expand the exponentials - quoting an identity scores nothing there. When squaring to find sinh x from cosh x, use the given sign of x to pick the root. Several of the identities above are printed in the formula booklet, so check it - but fluency in proving them is a standard proof mark that the booklet cannot give you.
Domains and ranges. sinh^-1 (also written arsinh): domain all real x. cosh^-1 (arcosh): x >= 1, because cosh never drops below 1 - the principal inverse takes y >= 0. tanh^-1 (artanh): |x| < 1, since tanh only takes values in (-1, 1). The Edexcel 9FM0 formula booklet quotes the logarithmic forms below; the derivations are, however, standard proof questions, so learn the method as well as the result.
The derivation method. Write y = f^-1(x), so x = f(y); replace the hyperbolic function by exponentials and solve the resulting quadratic in e^y.
Worked example (arsinh). Prove sinh^-1 x = ln(x + root(x^2 + 1)). Let y = sinh^-1 x, so x = (e^y - e^-y)/2. Multiply both sides by 2e^y: 2x e^y = e^(2y) - 1, so e^(2y) - 2x e^y - 1 = 0. Quadratic formula in e^y: e^y = x ± root(x^2 + 1). Since root(x^2 + 1) > |x| for every real x, the minus branch is always negative and rejected (e^y > 0), so e^y = x + root(x^2 + 1) > 0 for all x. Hence y = ln(x + root(x^2 + 1)).
The other two forms. cosh^-1 x = ln(x + root(x^2 - 1)) for x >= 1: both branches are positive here, but x - root(x^2 - 1) = 1/(x + root(x^2 - 1)) is at most 1, and the principal value needs e^y = cosh y + sinh y >= 1 (using y >= 0), so take the plus root; check x = 1 gives ln 1 = 0 ✓. tanh^-1 x = (1/2) ln((1 + x)/(1 - x)) for |x| < 1: from x = tanh y = (e^(2y) - 1)/(e^(2y) + 1), rearrange x(e^(2y) + 1) = e^(2y) - 1 to e^(2y) = (1 + x)/(1 - x); the fraction is positive exactly when |x| < 1, which is where the domain comes from. Derivatives, by implicit differentiation or the chain rule: d/dx sinh^-1 x = 1/root(1 + x^2), d/dx cosh^-1 x = 1/root(x^2 - 1) for x > 1, d/dx tanh^-1 x = 1/(1 - x^2) for |x| < 1.
Quoting the domain restriction earns marks even when the question forgets to ask for it. The sentence rejecting the rejected root (e^y > 0, or e^y >= 1 for arcosh) is the A-grade discriminator in these proofs. Remember sinh^-1 is odd: sinh^-1(-x) = -sinh^-1 x, while cosh^-1 is not. For x < 1 values plugged into cosh^-1 there is no real answer - state the domain, do not compute.
Differentiation. From the definitions: d/dx sinh x = cosh x; d/dx cosh x = sinh x - no sign flip, unlike trig differentiation, the most common slip; d/dx tanh x = sech^2 x = 1 - tanh^2 x. Chain rule versions appear constantly: d/dx sinh(kx) = k cosh(kx), d/dx cosh(x/a) = (1/a) sinh(x/a). The inverse derivatives from Section 2 combine with the chain rule to give standard integrals: d/dx sinh^-1(x/a) = 1/root(x^2 + a^2) and d/dx cosh^-1(x/a) = 1/root(x^2 - a^2) for x > a.
Integration. Reverse the derivatives: integral of sinh x dx = cosh x + c; integral of cosh x dx = sinh x + c; integral of tanh x dx = ln(cosh x) + c (substitute u = cosh x). Standard results: integral of 1/root(a^2 + x^2) dx = sinh^-1(x/a) + c = ln(x + root(x^2 + a^2)) + c; integral of 1/root(x^2 - a^2) dx = cosh^-1(x/a) + c for x > a. For squared functions use the half-identities: integral of sinh^2 x dx = sinh 2x/4 - x/2 + c and integral of cosh^2 x dx = sinh 2x/4 + x/2 + c.
Arc length and the catenary. Core Pure 2 requires the arc length of a curve y = f(x): L = integral of root(1 + (dy/dx)^2) dx between the limits. The catenary y = a cosh(x/a) is the shape of a heavy cable hanging under its own weight, and it is the classic arc length question: dy/dx = sinh(x/a), so 1 + (dy/dx)^2 = 1 + sinh^2(x/a) = cosh^2(x/a) by cosh^2 - sinh^2 = 1, and the square root disappears exactly: L = integral of cosh(x/a) dx = a sinh(x/a) evaluated between limits. Worked example: y = 3 cosh(x/3) from x = 0 to x = 3: L = integral from 0 to 3 of cosh(x/3) dx = [3 sinh(x/3)] from 0 to 3 = 3 sinh 1 ≈ 3.53. Endpoints: y(0) = 3 is the lowest point; y(3) = 3 cosh 1 ≈ 4.63.
The catenary collapse (1 + sinh^2 = cosh^2) is exactly why examiners choose this curve - show the identity step explicitly for the method mark. Quote the exact form (a sinh 1) before decimalising, to 3 significant figures only when asked. When a chain spans -b <= x <= b, use even symmetry and double the half-length from 0 to b. For arc lengths of other curves, complete the square inside the root to reach the standard integral forms above.