<< Chapter < Page Chapter >> Page >
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 andhyperbolic functions, give the following “addition formulas:”

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

sin ( z + w ) = sin ( z ) cos ( w ) + cos ( z ) sin ( w ) .
cos ( z + w ) = cos ( z ) cos ( w ) - sin ( z ) sin ( w ) .
sinh ( z + w ) = sinh ( z ) cosh ( w ) + cosh ( z ) sinh ( w ) .
cosh ( z + w ) = cosh ( z ) cosh ( w ) + sinh ( z ) sinh ( w ) .

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

First, for any two real numbers x and 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 ) ,

which, equating real and imaginary parts, gives that

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

and

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

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

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

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

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

  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:
    sin 2 ( z ) = 1 - cos ( 2 z ) 2 ,
    and
    cos 2 ( z ) = 1 + cos ( 2 z ) 2 .

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

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

0 = sin 2 ( π ) = 1 - cos ( 2 π ) 2

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

The periodicity of the cosine function is proved similarly.

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

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

HINT: If z = a + b i , then z = r ( a r + b r i . Observe that - 1 a r 1 , - 1 b r 1 , and ( a r ) 2 + ( b r ) 2 = 1 . Show that there exists a unique 0 θ < 2 π such that a r = cos θ and b r = sin θ .

Get the best College algebra course in your pocket!





Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Analysis of functions of a single variable' conversation and receive update notifications?

Ask