The Relations Between the Circular and the Hyperbolic Functions
Chapter Thirty-One
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 70 to 71 of 303
In one line
Put an imaginary angle into a circular function and a hyperbolic function comes out, and the other way round.
The six relations
MU's label is "Relations between circular and hyperbolic functions". These are what she means, and they are set as a group.
cos(i x) = cosh(x)
sin(i x) = i sinh(x)
tan(i x) = i tanh(x)
cosh(i x) = cos(x)
sinh(i x) = i sin(x)
tanh(i x) = i tan(x)
Read the pattern. The two even functions, cos and cosh, swap into each other with no i attached. The two odd ones pick up a factor of i. The relations are perfectly symmetric: each of the six is the statement above it read the other way.
Why they hold
Both families are built out of the same exponential, and the only difference is where the i sits.
cos(z) = (e^(i z) + e^(-i z))/2
cosh(z) = (e^z + e^(-z))/2
Put z = i x into the first one. Then iz is i squared x, which is minus x, and minus iz is plus x.
cos(i x) = (e^(-x) + e^x)/2
(e^(-x) + e^x)/2 = cosh(x)
That is the first relation, in two lines. The sine goes the same way, and the i comes out because the definition of the sine carries a division by 2i.
sin(i x) = (e^(-x) - e^x)/(2i)
(e^(-x) - e^x)/(2i) = i sinh(x)
To see that last step, multiply top and bottom by i: the denominator becomes minus 2 and the numerator becomes i(e to the minus x minus e to the x), so the whole thing is i times (e to the x minus e to the minus x) over 2, which is i sinh x.
What the relations are for
Three uses, all examined.
Separating a function of x + iy into its parts. The chapter on that used exactly these relations at step three, and could not have been written without them.
Converting an identity. Any circular identity becomes a hyperbolic one, and the next chapter turns that into a rule.
Evaluating a function at a complex point. A question asking for the value of cosh(2 + 3i) or sin(1 + i) is answered with the compound-angle formula and these relations.
Worked: find the value of tan(i), in terms of a hyperbolic function.
tan(i) = i tanh(1)
Worked: show that cosh(i pi) is minus one.
cosh(i pi) = cos(pi)
cos(pi) = -1
So the hyperbolic cosine, which for real arguments is never less than one, takes the value minus one at an imaginary argument. That is worth noticing: the bound "cosh is at least one" is a fact about real arguments only, exactly as "cos is at most one" is.
The Relations Between the Circular and the Hyperbolic Functions
The one that catches people
The relations pick up an i, and the i is easy to drop. Here is the difference it makes.
sin(i) = i sinh(1)
The sine of i is purely imaginary, with modulus sinh 1, which is about 1.175. Writing sin(i) = sinh(1) loses the i and gives a real number, which is wrong, and the same slip in the middle of a longer question turns a real answer imaginary or the other way round.
The way to keep it straight: the sine is an odd function and so is sinh, and an odd function of an imaginary argument has to be imaginary, because sin(minus z) is minus sin(z). The cosine is even, so the cosine of an imaginary argument is real. Parity settles where the i goes, every time.
The periods, which are the other side of the same coin
The relations say something surprising about the hyperbolic functions: they are periodic, with an imaginary period.
cosh(x + 2 i pi) = cosh(x)
sinh(x + 2 i pi) = sinh(x)
For real arguments cosh and sinh never repeat a value, so periodicity looks absurd. But cosh(z) is cos(iz), and the cosine has period 2 pi, so cosh has period 2 pi i. The two families are the same functions looked at along two perpendicular directions.
Check yourself
cos(2i) = cosh(2)
sin(3i) = i sinh(3)
cosh(i pi/2) = 0
tanh(i pi/4) = i
sin(i)^2 = -sinh(1)^2
cos(i)^2 + sin(i)^2 = 1
The third line is a good test: cosh(i pi over 2) is cos(pi over 2), which is zero, so the hyperbolic cosine does have a zero after all, at an imaginary argument. The fifth line shows the i squared doing its work, and the last shows that the fundamental circular identity holds at a complex argument too, as the earlier chapter said it would.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.