As a temporary diversion, look at differentiating eˣ+e⁻ˣ. Not hard; we get eˣ-e⁻ˣ. The sum of these results is 2eˣ and the difference is 2e⁻ˣ. Consider instead eⁱˣ+e⁻ⁱˣ; ignoring the meaning of the multiplier i we get eⁱˣ-e⁻ⁱˣ. Similar to what we just did we now have a sum of 2eⁱˣ and a difference of 2e⁻ⁱˣ. It is already tempting to name two pairs of functions, ½(eⁱˣ-e⁻ⁱˣ) & ½(eⁱˣ+e⁻ⁱˣ), ½(eˣ-e⁻ˣ) & ½(eˣ+e⁻ˣ) and sure enough, we would not be the first to do that. I’m going to call the first pair f and fc, because they are complements, and similarly the second pair g and gc.
Now look at what happens if we square one of them.
fc(x)² = [½(eⁱˣ+e⁻ⁱˣ)]² = ¼(e²ⁱˣ+2 + e⁻²ⁱˣ)= ¼(e²ⁱˣ+e⁻²ⁱˣ)+ ½ = ½ (f(2x) +1)
which I will rewrite as 2f²(x) - 1 = f(2x). Intriguing.
More intriguing still, look at the product of the complementary pair, f. fc
f.fc = ½(eⁱˣ+e⁻ⁱˣ) . ½(eˣ-e⁻ˣ) = ½ ½(e²ⁱˣ-e⁻²ⁱˣ) = ½ f(2x) .
So 2 f(x) fc(x) = fc(2x).
That is, I hope, familiar to you, as a functional mapping you have seen before.
Exercise:
9. Show that fc(2x) = fc²(x) - f²(x) = 1 - 2 f²(x)
10. Show that fc²(x) + f²(x) = 1
Hopefully you now recognise these as behaving exactly as sine and cosine. How could we prove they are the same thing? What do you need to prove to establish complete equivalence? Function f is sine and fc is cosine.
What about g and gc? Here’s some simple (now, they’re simple) things to prove:
11. gc²(x) - g²(x) = 1
12. g(2x) = 2 g(x) gc(x)
13. gc(2x) = gc²(x) - 1 = 2 g²(x) + 1 = gc²(x) + g²(x)
These behave very much like the f and fc pair, but the g switches sign every time it is squared – that’s the effect of the missing i².
The pair in f are sin and cos; the pair in g are sinh and cosh, the hyperbolic versions, pronounced shine and cosh (and the tangent equivalent, tanh, is prounced th-an, more as thatch and not as than). Similalrly, we have sech, cosech and coth. So it might have been better to learn of sinh and cosh first and to adjust for the i²=-1 in handling sine and cosine.
Declaring some of these related identities, which of course you should prove:
cosh²A - sinh²A = 1;
sinh (A+B) = sinhAcoshB+coshAsinhB;
cosh (A+B) = coshAcoshB+sinhAsinhB.
Exercises:
14. Discover whether sinh and cosh are odd or even functions. Compare this with sin and cos.
15. Write down sin 3A only in terms of sinA. Predict what sinh 3A will be and then discover that.
16. Prove that sech² A = 1 – tanh² A (the sinh² on the top forces a change of sign)
17. Prove that cosech²A = coth²A - 1
14. sinh is an odd function , cosh is even, exactly matching sin and cos.
15. sin 3A = 3sinA – 4sin3A; sinh 3A = 3sin A + 4sinh3A
The definition of eˣ is that it is the function which is the same as its differential. That results in the expansion eˣ = 1+x+x²/2! +x³/3! +x⁴/4!+….
Similarly e⁻ˣ = 1-x+x²/2! -x³/3! +x⁴/4!-….
So cosh x = 1 + x2/2! + x4/4!+….
And sinh x = x + x3/3! +x5/5!-….
The graph of y = sinh x is approximately like a positive cubic and y = cosh x looks approximately like a positive quadratic.
The inverse function sinh⁻¹ x or arsinh x has an identity in logs.
Let y=arsinh x. Then x =sinhy = ½ (eʸ – e⁻ʸ).
So eʸ=2x+e⁻ʸ and, multiplying again by eʸ, e²ʸ= 2xeʸ - 1
This is a quadratic e²ʸ - 2xeʸ + 1 = 0 such that
eʸ = x ± √ (x²+1).
Hence y = arsinh x = ln( x ± √ (x²+1)).
You should satisfy yourself that rejecting the negative root is justified.
18. Show that arcosh x = ln( x ± √ (x² - 1).) and state the range of values for which it is valid.
19. Show that artanh x = ½ ln((1+x)/(1-x)) and state the range of values for which it is valid.
Problems to solve (this is FP3 in Edexcel) would include expressing arsinh² in logs and solving
a cosh² A + b sinh A = c, giving answers in logs — many problems equivalent to those in C1- C4 transfer quite well.
18 x>=0 19 abs(x) <1
Here, at least, is some differentiation to do:
20. Show that d/dx (cosech x) = -cothx cosechx
End of hyperbolic functions 1
