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 Math Help > Trigonometry > Trig Equivalences > Hyperbolic Functions

# Hyperbolic Functions

cosh x, pronounced the way it looks, is the hyperbolic cosine of x, (ex+e-x)/2

sinh x, sometimes pronounced "shine x", is the hyperbolic sine of x, (ex-e-x)/2.

The "circular functions", sin and cos, are so called because (cos t, sin t) lies on the circle x² + y² = 1, whereas the hyperbolic ordered pair
(cosh t, sinh t) lies on the hyperbola x² - y² = 1.

Together with the exponential function exp(x), which is defined as ex, these circular and hyperbolic functions form a nice, happy family.  Let me show you how, starting from the power series for ex, cos(x) and sin(x):

### cos(x) and sin(x) and their relation to exp(x)

exp(x) = ex = 1 + x/1! + x2/2! + x3/3! + ...

cos(x) = 1  - x2/2! + x4/4! - x6/6! + ...

sin(x) = x/1! - x3/3! + x5/5! - ...

From the series expansion of exp(x), we can see that exp(ix) is given by

exp(ix) = eix = 1 + ix/1! + i2x2/2! + i3x3/3! + i4x4/4! + ...

exp(ix) = eix = 1 + ix/1! - x2/2! - ix3/3! + x4/4! + ...

Collecting the real terms together and the imaginary terms together, we get

exp(ix) = eix = cos(x) + i sin(x)

Now replacing x with -ix, we get the equivalent identity,

exp(x) = ex = cos(-ix) + i sin(-ix)

exp(x) = ex = cos(ix) - i sin(ix)

### cosh(x) and sinh(x) and their relation to exp(x):

Remember the definitions of cosh and sinh:

cosh(x) = (ex+e-x)/2.

sinh(x) = (ex-e-x)/2

So their sum cosh(x)+sinh(x) = ex,
and their difference cosh(x)-sinh(x) = e-x.

### cosh(x) and sinh(x) and their relation to cos(ix) and sin(ix):

cosh(ix) = (eix+e-ix)/2 =

(cos(x) + i sin(x) + cos(-x) + i sin(-x))/2 =

(cos(x) + i sin(x) + cos(x) - i sin(x))/2 =

cos(x)

sinh(ix) = (eix-e-ix)/2 =

(cos(x) + i sin(x) - cos(-x) - i sin(-x))/2 =

(cos(x) + i sin(x) - cos(x) + i sin(x))/2 =

i sin(x)

Since cosh(ix) = cos(x), it follows that cosh(x) = cos(-ix) = cos(ix)

Since sinh(ix) = i sin(x), it follows that sinh(x) = i sin(-ix) = -i sin(ix)

So cos(ix) = cosh(x)
and sin(ix) = i sinh(x)

### arcsinh(x) = ln(x + sqrt(x² + 1))

The proof of this hyperbolic equivalence, and the one that follows it, is done by starting with the result, and working backwards.  That is, I wrote "x = sinh(ln(x + sqrt(x²+1)))" on the bottom of the page, then above it I wrote "x = exp(ln(...", and then kept writing each simplification on the line above that, until I reached the top line, which is a self-evident equivalence.  I used this technique to prove the quadratic formula, too.

Proof:

2x² + 2x sqrt(x²+1) = x² + 2x sqrt(x²+1) + (x²+1) - 1
2x(x+sqrt(x²+1)) = (x+sqrt(x²+1))² - 1
2x = x+sqrt(x²+1) - 1/(x+sqrt(x²+1))
x = exp(ln(x+sqrt(x²+1)))/2 - exp(-ln(x+sqrt(x²+1)))/2
x = sinh(ln(x + sqrt(x²+1)))

### arccosh(x) = ln(x + sqrt(x² - 1))

Proof -- very similar:

2x² + 2x sqrt(x²-1) = x² + 2x sqrt(x²-1) + (x²-1) + 1
2x(x+sqrt(x²-1)) = (x+sqrt(x²-1))² + 1
2x = x + sqrt(x²-1) + 1/(x+sqrt(x²-1))
x = exp(ln(x + sqrt(x²-1))) + exp(-ln(x + sqrt(x²-1)))
x = cosh(ln(x + sqrt(x²-1)))

Note: you can prove just as easily that arccosh(x) = ln(x - sqrt(x² - 1));  You should be able to see that

x - sqrt(x²-1)

is the reciprocal of

x + sqrt(x²-1)

so the natural log of one is the negative of the natural log of the other, and since cosh(y) = cosh(-y), the two arccosh's of x are negatives of one another.  However, it is customary to take ln(x + sqrt(x²-1)) as the principle arccosh of x.

### arctanh(x) = (1/2) (ln(1+x) - ln(1-x))

Proof:

x = 2x / 2
x = ((1+x)-(1-x)) / ((1+x)+(1-x))
x = ((1+x)sqrt(1/(1-x²))-(1-x)sqrt(1/(1-x²))) / ((1+x)sqrt(1/(1-x²))+(1-x)sqrt(1/(1-x²)))
x = (sqrt((1+x)²/(1-x²))-sqrt((1-x)²/(1-x²))) / (sqrt((1+x)²/(1-x²))+sqrt((1-x)²/(1-x²)))
x = (sqrt((1+x)/(1-x))-sqrt((1-x)/(1+x))) / (sqrt((1+x)/(1-x))+sqrt((1-x)/(1+x)))
x = (exp(ln sqrt((1+x)/(1-x)))-exp(-ln sqrt((1+x)/(1-x)))) / (exp(ln sqrt((1+x)/(1-x)))+exp(-ln sqrt((1+x)/(1-x))))
x = tanh(ln sqrt((1+x)/(1-x)))
x = tanh((1/2) (ln(1+x) - ln(1-x)))

### Summary

Series expansions:

exp(x) = 1 + x/1! + x2/2! + x3/3! + x4/4! + ...
cos(x) = 1 - x2/2! + x4/4! - x6/6! + ...
sin(x) = x/1! - x3/3! + x5/5! - x7/7! + ...
cosh(x) = 1 + x2/2! + x4/4! + x6/6! + ...
sinh(x) = x/1! + x3/3! + x5/5! + x7/7! + ...

Definitions of cosh, sinh, tanh in terms of exp:

cosh(x) = (ex+e-x)/2
sinh(x) = (ex-e-x)/2
tanh(x) = (ex-e-x)/(ex+e-x)

Equivalences:

exp(x) = cosh(x) + sinh(x)
exp(ix) = cos(x) + i sin(x)

cos(ix) = cosh(x)
sin(ix) = i sinh(x)

cos²(x) + sin²(x) = 1
cosh²(x) - sinh²(x) = 1

arsinh(x) = ln(x + sqrt(x² + 1))
arcosh(x) = ln(x + sqrt(x² - 1))   or   ln(x - sqrt(x² - 1))
arctanh(x) = (1/2) (ln(1+x) - ln(1-x))

### Related Pages in this website

Summary of trig identities: Euler identities -- based on eix=cos(x)+i sin(x), such as cosh(ix) = cos(x), etc.

Hypergeometric Function -- the general form of various functions, expressed in terms of their power series

Euler and Exponents -- Euler's formula, and xy

The Quadratic Formula -- another proof by working backwards

The webmaster and author of this Math Help site is Graeme McRae.