Hyperbolic Identities for JEE
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Direct answer
cosh x and sinh x are the even and odd halves of e^x — (e^x + e^(−x))/2 and (e^x − e^(−x))/2 — and once that split is visible, every hyperbolic identity becomes a two-line exercise in exponents rather than a memory task. The master relation is cosh²x − sinh²x = 1, with the minus sign where circular functions carry a plus. Addition theorems mirror trigonometry: sinh(x ± y) = sinh x cosh y ± cosh x sinh y and cosh(x ± y) = cosh x cosh y ± sinh x sinh y. Ranges decide equations: sinh and tanh take restricted sets (tanh stays inside (−1, 1)) while cosh ≥ 1, so cosh x = 1/2 has no real solution.
What you must remember
- Definitions and inversion: e^x = cosh x + sinh x and e^(−x) = cosh x − sinh x; cosh 0 = 1, sinh 0 = 0; sinh is odd, cosh is even.
- Pythagorean set: cosh²x − sinh²x = 1; 1 − tanh²x = sech²x; coth²x − 1 = csch²x — every sign flipped from its circular analogue.
- Double arguments: sinh 2x = 2 sinh x cosh x; cosh 2x = cosh²x + sinh²x = 2cosh²x − 1 = 1 + 2sinh²x.
- Addition consequences: cosh x cosh y ± sinh x sinh y = cosh(x ± y), and sinh(x + y)sinh(x − y) = sinh²x − sinh²y.
- Ranges for equations: sinh x ∈ ℝ, cosh x ∈ [1, ∞), tanh x ∈ (−1, 1) — so tanh x = 2 is impossible while sinh x = 2 is fine.
- Calculus anchors: d/dx sinh x = cosh x, d/dx cosh x = sinh x (no minus sign), d/dx tanh x = sech²x, and ∫sech²x dx = tanh x.
- Complex bridge: sin(ix) = i sinh x and cos(ix) = cosh x — Osborne's rule in one glance: products of two sines flip sign when converting trig identities to hyperbolic ones.
Working with the definitions
Three short computations cover most asked forms. First, if sinh x = 3/4, find tanh x: the master identity gives cosh²x = 1 + 9/16 = 25/16, and cosh is always positive, so cosh x = 5/4 and tanh x = (3/4)/(5/4) = 3/5 — the 3-4-5 triangle reappearing with one sign changed. Second, evaluate cosh(2 ln 2) straight from the definition: e^(2 ln 2) = 4 and e^(−2 ln 2) = 1/4, so cosh = (4 + 1/4)/2 = 17/8. Third, solve sinh x = 2: use the inversion e^x = sinh x + cosh x = 2 + √5, so x = ln(2 + √5). That inversion formula is the equation-solving tool of the chapter — every sinh or cosh equation reduces to an exponential one through it.
Contrast the failure case: tanh x = 2 has no real solution because |tanh x| < 1 always; recognising impossibility from the range is itself a tested skill, typically as a "number of real solutions" item.
Where hyperbolic questions appear
Recent Main papers ask these sparingly; the identities surface mainly in Advanced-level calculus — derivatives chained with the chain rule, integrals of sech²x — and in older papers' algebra of e^x, so treat the chapter as identity fluency rather than a question factory. Three traps recur. Writing cosh²x + sinh²x = 1 as if it were the Pythagorean identity (that sum is cosh 2x; the sign flip is the entire subject). Assuming cosh is bounded like cosine — it is not, growing like e^x/2. And inserting a spurious ± when extracting cosh from sinh: cosh x = +√(1 + sinh²x) with no alternative, because cosh never dips below 1, which is precisely why the 3-4-5 computation above needed no sign discussion.
Frequently asked questions
What is the hyperbolic Pythagorean identity?
cosh²x − sinh²x = 1, with companions 1 − tanh²x = sech²x and coth²x − 1 = csch²x — all signs flipped from the circular versions.
If sinh x = 3/4, what is tanh x?
cosh x = √(1 + 9/16) = 5/4, taken positive because cosh ≥ 1, so tanh x = 3/5.
Does cosh x = 1/2 have a real solution?
No — cosh x = (e^x + e^(−x))/2 ≥ 1 for all real x by AM-GM, so the equation is impossible.
How do I solve sinh x = k?
Use e^x = sinh x + cosh x = k + √(1 + k²), giving x = ln(k + √(1 + k²)).
Which derivative facts matter most?
d/dx sinh x = cosh x, d/dx cosh x = sinh x with no minus sign, and d/dx tanh x = sech²x — the missing minus is a favourite option trap.