From the definitions

coshx = 1 2 (ex + e-x) sinhx = 1 2 (ex - e-x)

we see that coshx is always positive, it is an even function and its derivative is sinhx and sinhx is zero when x=0 so the graph of coshx is U shaped with a minimum at (0,1), as this is the only turning point and coshx increases as x increases for x > 0.

The derivative of sinhx is coshx so the graph of sinhx is everywhere increasing with the only turning point a point of inflexion at the origin. As sinh is an odd function the graph has rotational symmetry about the origin. Clearly for all x we have sinhx < coshx so the graph of sinh is below the graph of cosh and they get closer as x increases.

1. First, I have to prove that: cosh² x - sinh² x = 1. From the definition coshx + sinhx = e^x and coshx - sinhx = e^-x so

cosh2 x - sinh2 x = (coshx + sinhx)(coshx - sinhx) = ex ×e-x = 1.

2. In the second part, I have to prove that: sinh2x = 2sinhx coshx.

I know that:

sinh2x = 1 2 (e2x - e-2x)

and

2sinhx coshx = 1 2 (ex - e-x)(ex + e-x)= 1 2 (e2x-e-2x).

The two results above are equal, so I have proved that: sinh2x = 2sinhx coshx.

3. I have to prove that:sinh(n+1)x = sinhnx coshx + coshnx sinhx,

Starting again from the definition, I can write the following expression for the terms:

sinhnx coshx + coshnx sinhx = 1 4 [(enx - e-nx)(ex + e-x)+ (enx + e-nx)(ex - e-x)] = 1 4 [2e(n+1)x-2e-(n+1)x] = 1 2 [e(n+1)x - e-(n+1)x] = sinh(n+1)x .

4. I have to prove the following inequality: sinhnx ³ n sinhx.

This could be demonstrated using induction. It is clear that for n = 1 I obtain an equality, so I look to prove the inequality for n = 2. Then I shall consider it true for n=k and prove it for (k+1).

As coshx > 0 for all x it follows that sinh 2x = 2sinhx coshx > 2sinhx for all x so the statement is true for n=1 and n=2. Now assume the statement is true for n=k and consider sinh(k+1)x. We have, as cosh is positive for all x,

sinh(k+1)x = sinhkx coshx + coshkx sinhx ³ sinhkx + sinhx

and so if sinhkx ³ ksinhx then

sinh(k+1)x ³ ksinhx + sinhx = (k+1)sinhx

and hence by the axiom of mathematical induction the statement is true for all positive integers n.