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Textbook by: Lawrence Baggett. E-mail the author

# The Trigonometric and Hyperbolic Functions

Module by: Lawrence Baggett. E-mail the author

Summary: Two theorems covering differentiation of trigonometric and hyperbolic functions, including practice exercises corresponding to the theorems.

The laws of exponents and the algebraic connections between the exponential function and the trigonometric and hyperbolic functions, give the following “addition formulas:”

## Theorem 1

The following identities hold for all complex numbers zz and w.w.

sin ( z + w ) = sin ( z ) cos ( w ) + cos ( z ) sin ( w ) . sin ( z + w ) = sin ( z ) cos ( w ) + cos ( z ) sin ( w ) .
(1)
cos ( z + w ) = cos ( z ) cos ( w ) - sin ( z ) sin ( w ) . cos ( z + w ) = cos ( z ) cos ( w ) - sin ( z ) sin ( w ) .
(2)
sinh ( z + w ) = sinh ( z ) cosh ( w ) + cosh ( z ) sinh ( w ) . sinh ( z + w ) = sinh ( z ) cosh ( w ) + cosh ( z ) sinh ( w ) .
(3)
cosh ( z + w ) = cosh ( z ) cosh ( w ) + sinh ( z ) sinh ( w ) . cosh ( z + w ) = cosh ( z ) cosh ( w ) + sinh ( z ) sinh ( w ) .
(4)

### Proof

We derive the first formula and leave the others to an exercise.

First, for any two real numbers xx and y,y, we have

cos ( x + y ) + i sin ( x + y ) = e i ( x + y ) = e i x e i y = ( cos x + i sin x ) × ( cos y + i sin y ) = cos x cos y - sin x sin y + i ( cos x sin y + sin x cos y ) , cos ( x + y ) + i sin ( x + y ) = e i ( x + y ) = e i x e i y = ( cos x + i sin x ) × ( cos y + i sin y ) = cos x cos y - sin x sin y + i ( cos x sin y + sin x cos y ) ,
(5)

which, equating real and imaginary parts, gives that

cos ( x + y ) = cos x cos y - sin x sin y cos ( x + y ) = cos x cos y - sin x sin y
(6)

and

sin ( x + y ) = sin x cos y + cos x sin y . sin ( x + y ) = sin x cos y + cos x sin y .
(7)

The second of these equations is exactly what we want, but this calculation only shows that it holds for real numbers xx and y.y. We can use the Identity Theorem to show that in fact this formula holds for all complex numbers zz and w.w. Thus, fix a real number y.y. Let f(z)=sinzcosy+coszsiny,f(z)=sinzcosy+coszsiny, and let

g ( z ) = sin ( z + y ) = 1 2 i ( e i ( z + y ) - e - i ( z + y ) = 1 2 i ( e i z e i y - e - i z e - i y ) . g ( z ) = sin ( z + y ) = 1 2 i ( e i ( z + y ) - e - i ( z + y ) = 1 2 i ( e i z e i y - e - i z e - i y ) .
(8)

Then both ff and gg are power series functions of the variable z.z. Furthermore, by the previous calculation, f(1/k)=g(1/k)f(1/k)=g(1/k) for all positive integers k.k. Hence, by the Identity Theorem, f(z)=g(z)f(z)=g(z) for all complex z.z. Hence we have the formula we want for all complex numbers zz and all real numbers y.y.

To finish the proof, we do the same trick one more time. Fix a complex number z.z. Let f(w)=sinzcosw+coszsinw,f(w)=sinzcosw+coszsinw, and let

g ( w ) = sin ( z + w ) = 1 2 i ( e i ( z + w ) - e - i ( z + w ) = 1 2 i ( e i z e i w - e - i z e - i w ) . g ( w ) = sin ( z + w ) = 1 2 i ( e i ( z + w ) - e - i ( z + w ) = 1 2 i ( e i z e i w - e - i z e - i w ) .
(9)

Again, both ff and gg are power series functions of the variable w,w, and they agree on the sequence {1/k}.{1/k}. Hence they agree everywhere, and this completes the proof of the first addition formula.

## Exercise 1

1. Derive the remaining three addition formulas of the preceding theorem.
2. From the addition formulas, derive the two “half angle” formulas for the trigonometric functions:
sin2(z)=1-cos(2z)2,sin2(z)=1-cos(2z)2,
(10)
and
cos2(z)=1+cos(2z)2.cos2(z)=1+cos(2z)2.
(11)

## Theorem 2

The trigonometric functions sinsin and coscos are periodic with period 2π;2π; i.e., sin(z+2π)=sin(z)sin(z+2π)=sin(z) and cos(z+2π)=cos(z)cos(z+2π)=cos(z) for all complex numbers z.z.

### Proof

We have from the preceding exercise that sin(z+2π)=sin(z)cos(2π)+cos(z)sin(2π),sin(z+2π)=sin(z)cos(2π)+cos(z)sin(2π), so that the periodicity assertion for the sine function will follow if we show that cos(2π)=1cos(2π)=1 and sin(2π)=0.sin(2π)=0. From part (b) of the preceding exercise, we have that

0 = sin 2 ( π ) = 1 - cos ( 2 π ) 2 0 = sin 2 ( π ) = 1 - cos ( 2 π ) 2
(12)

which shows that cos(2π)=1.cos(2π)=1. Since cos2+sin2=1,cos2+sin2=1, it then follows that sin(2π)=0.sin(2π)=0.

The periodicity of the cosine function is proved similarly.

## Exercise 2

1. Prove that the hyperbolic functions sinhsinh and coshcosh are periodic. What is the period?
2. Prove that the hyperbolic cosine cosh(x)cosh(x) is never 0 for xx a real number, that the hyperbolic tangent tanh(x)=sinh(x)/cosh(x)tanh(x)=sinh(x)/cosh(x) is bounded and increasing from RR onto (-1,1),(-1,1), and that the inverse hyperbolic tangent has derivative given by tanh-1'(y)=1/(1-y2).tanh-1'(y)=1/(1-y2).
3. Verify that for all y(-1,1)y(-1,1)
tanh-1(y)=ln(1+y1-y).tanh-1(y)=ln(1+y1-y).
(13)

## Exercise 3: Polar coordinates

Let zz be a nonzero complex number. Prove that there exists a unique real number 0θ<2π0θ<2π such that z=reiθ,z=reiθ, where r=|z|.r=|z|.

HINT: If z=a+bi,z=a+bi, then z=r(ar+bri.z=r(ar+bri. Observe that -1ar1,-1ar1,-1br1,-1br1, and (ar)2+(br)2=1.(ar)2+(br)2=1. Show that there exists a unique 0θ<2π0θ<2π such that ar=cosθar=cosθ and br=sinθ.br=sinθ.

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