Relationships between exponential function and trigonometric functions

A similar function is the exponential function ${\displaystyle e(\theta )\,}$ which is defined by the statement: the rate of change of ${\displaystyle e(\theta )\,}$ is ${\displaystyle e(\theta )\,}$ and the value of ${\displaystyle e(0)\,}$ is 1. Here we only have to apply the rate of change operator once to get back where we started. Explicitly:

 ${\displaystyle {\frac {e(\theta +\delta \theta )-e(\theta )}{\delta \theta }}}$ tends to ${\displaystyle e(\theta )\,}$ as ${\displaystyle \delta \theta \,}$ tends to zero.


which can be rewritten replacing ${\displaystyle \theta \,}$ by ${\displaystyle i\theta \,}$ where ${\displaystyle \,i}$ is any constant number, to get:

 ${\displaystyle {\frac {e(i\theta +\delta i\theta )-e(i\theta )}{\delta i\theta }}}$ tends to ${\displaystyle e(i\theta )\,}$ as ${\displaystyle \delta \theta \,}$ tends to zero.


because ${\displaystyle i\,}$ is a constant, ${\displaystyle \delta i\theta =i\delta \theta \,}$, performing this substitution, we get:

    ${\displaystyle {\frac {e(i\theta +i\delta \theta )-e(i\theta )}{i\delta \theta }}}$ tends to ${\displaystyle e(i\theta )\,}$ as ${\displaystyle \delta \theta \,}$ tends to zero.
${\displaystyle \Rightarrow {\frac {e(1(\theta +\delta \theta ))-e(i\theta )}{i\delta \theta }}}$ tends to ${\displaystyle e(i\theta )\,}$ as ${\displaystyle \delta \theta \,}$ tends to zero.
${\displaystyle \Rightarrow {\frac {de(i\theta )}{d\theta }}=i*e(i\theta )}$


That is, the rate of change of ${\displaystyle e(i\theta )\,}$ with ${\displaystyle \theta \,}$ is ${\displaystyle i*e(i\theta )\,}$. We can continue this process to find the rate of change of ${\displaystyle i*e(i\theta )\,}$ with ${\displaystyle \theta \,}$:

  ${\displaystyle {\frac {di*e(i\theta )}{d\theta }}}$ is the limit of:
${\displaystyle {\frac {i*e(i(\theta +\delta \theta ))-i*e(i\theta )}{\delta \theta }}}$ as ${\displaystyle \delta \theta \,}$ tends to zero, which is the same as:
${\displaystyle {\frac {i(e(i(\theta +\delta \theta ))-e(i\theta ))}{\delta \theta }}}$ as ${\displaystyle \delta \theta \,}$ tends to zero, which is:

   ${\displaystyle i{\frac {de(i\theta )}{d\theta }}=i*i*e(i\theta )}$


Performing this 4 time successively yields and comparing with the same action on the ${\displaystyle cos()\,}$ function:

   ${\displaystyle {\frac {de(i\theta )}{d\theta }}=i*e(i\theta )}$                         ${\displaystyle {\frac {dcos(\theta )}{d\theta }}=-sin(\theta )}$
${\displaystyle {\frac {di*e(i\theta )}{d\theta }}=i*i*e(i\theta )}$                     ${\displaystyle {\frac {d-sin(\theta )}{d\theta }}=-cos(\theta )}$
${\displaystyle {\frac {di*i*e(i\theta )}{d\theta }}=i*i*i*e(i\theta )}$                 ${\displaystyle {\frac {d-cos(\theta )}{d\theta }}=sin(\theta )}$
${\displaystyle {\frac {di*i*i*e(\theta )}{d\theta }}=i*i*i*i*e(i\theta )}$               ${\displaystyle {\frac {dsin(\theta )}{d\theta }}=cos(\theta )}$


If only there was a number ${\displaystyle i\,}$, such that ${\displaystyle i*i=-1}$, and hence ${\displaystyle i*i*i*i=-1*-1=1}$, then we could relate the function ${\displaystyle cos()\,}$ to the function ${\displaystyle e()\,}$. Fortunately, there is a number that will work: the square roots of -1. From here on ${\displaystyle i\,}$ will denote a square root of -1. ${\displaystyle -i*-i=(-1)*i*(-1)*i=(-1)*(-1)*i*i=(1)*(-1)=-1}$ is also a solution. We can expect then that ${\displaystyle cos(\theta )\,}$ is some linear combination of ${\displaystyle e(i\theta )\,}$ and ${\displaystyle e(-i\theta )\,}$, perhaps:

 ${\displaystyle cos(\theta )=A*e(i\theta )+B*e(-i\theta )\,}$


We know that ${\displaystyle cos(0)=1\,}$, so:

 ${\displaystyle cos(0)=1=A*e(i*0)+B*e(-i*0)=A*e(0)+B*e(0)=A+B\,}$


Finding the rate of change with ${\displaystyle \theta \,}$:

${\displaystyle \Rightarrow {\frac {dcos(\theta )}{d\theta }}={\frac {d({\frac {A*e(i\theta )}{d\theta }}+d(B*e(-i\theta ))}{d\theta }}}$
${\displaystyle =>-sin(\theta )=i*A*e(i\theta )+-i*B*e(-i\theta )=-cos({\frac {\pi }{2}}-\theta )}$


setting ${\displaystyle \theta =0\,}$, remembering that ${\displaystyle cos({\frac {\pi }{2}})=0\,}$,

${\displaystyle \Rightarrow 0=iA-iB=i(A-B)\Rightarrow A=B}$


So now we know that ${\displaystyle A+B=1}$ and ${\displaystyle A=B}$, so ${\displaystyle A+A=1}$, so ${\displaystyle A=1/2}$ and ${\displaystyle B=1/2}$.

To summarize what we know so far:

${\displaystyle cos(\theta )={\frac {e(i\theta )+e(-i\theta )}{2}}}$ where ${\displaystyle \theta \,}$ is in radians and ${\displaystyle i\,}$ is a square root of -1.


Replacing ${\displaystyle \theta \,}$ by ${\displaystyle -\theta \,}$ gives conversely:

${\displaystyle cos(-\theta )={\frac {e(-i\theta )+e(i\theta )}{2}}}$, the same formula, so we must have that ${\displaystyle cos(\theta )=cos(-\theta )\,}$.


Given:

${\displaystyle cos(\theta )={\frac {{\frac {e(i\theta )}{2}}+e(-i\theta )}{2}}}$


We can find the rate of change with θ of both sides to fined the sin() in terms of e()

    -sin(θ)  =  i+e( iθ)/2 - i*e(-iθ)/2

 =>  sin(θ)  =  i*e(-iθ)/2 - i*e( iθ)/2


Substituting -θ for θ gives:

     sin(-θ)  =  i*e(iθ)/2 - i*e(-iθ)/2 = -sin(θ)


thus sin() is an odd function, compare this to cos() which is an even function because cos(θ) = cos(-θ).

We can find the function e() in terms of cos() and sin():

 cos(θ) + i * sin(θ)  =  e(iθ)/2 + e(-iθ)/2 + i*i*e(-iθ)/2 - i*i*e(iθ)/2
=  e(iθ)/2 + e(-iθ)/2 -     e(-iθ)/2 +     e(iθ)/2
=  e(iθ)


This is called "Euler's Formula".

From the "Cosine Double Angle Formula", we know that:

cos(2θ) = 2 * cos(θ)**2 - 1


Let θ = π/2, so that cos(π/2) = 0, then:

   cos(2π/2) = 2 * cos(π/2)**2 - 1
=> cos (π)   = 2 * 0**2        - 1
=> cos (π)   = - 1


By Pythagoras:

   sin(π)**2 + cos(π)**2 = 1
=> sin(π)**2 + -1**2     = 1
=> sin(π)**2             = 0
=> sin(π)                = 0


Consequently, we can evaluate e(iπ) as:

 e(iπ) = cos(π) + i * sin(π) = -1 + i * 0 = -1;


Similarly, we can evaluate e(-iθ) from e(iθ):

   e( iθ) = cos( θ) + i * sin( θ)
=> e(-iθ) = cos(-θ) + i * sin(-θ)
=> e(-iθ) = cos(θ)  - i * sin( θ)  as cos() is even, sin() is odd



This result allows us to evaluate e(iθ)e(-iθ):

   e(iθ)e(-iθ) = (cos(θ) + i * sin(θ))(cos(θ) - i * sin(θ))
= (cos(θ)**2 - i*i*sin(θ)**2)
=  cos(θ)**2 +     sin(θ)**2         as i*i = -1
=  1                                 Pythagoras


Starting again from the "Cosine Double Angle Formula", we know that:

cos(2θ) = 2 * cos(θ)**2 - 1


Replace cos() by its formulation in e():

cos(θ) = e(iθ)/2 + e(-iθ)/2


to get:

e(2iθ)/2 + e(-2iθ)/2 = 2 * (e(iθ)/2 + e(-iθ)/2)**2                     - 1
= 2 * (e(iθ)/2)**2+(e(-iθ)/2)**2 +2e(iθ)e(-iθ)/4) - 1
=     (e(iθ)**2)/2+(e(-iθ)**2)/2 + e(iθ)e(-iθ)    - 1


But

e(iθ)e(-iθ) = 1


so we continue the algebraic simplification to get:

 e(2iθ)/2 + e(-2iθ)/2 =     (e(iθ)**2)/2+(e(-iθ)**2)/2 + e(iθ)e(-iθ)    - 1
=     (e(iθ)**2)/2+(e(-iθ)**2)/2 + 1              - 1
=     (e(iθ)**2)/2+(e(-iθ)**2)/2


Again

e(iθ)e(-iθ) = 1


so we are forced to conclude that

 e( 2iθ) = e( iθ)**2  and
e(-2iθ) = e(-iθ)**2   for any angle θ.


The e() function is behaving like exponentiation, that is we can write:

 e(iθ) = e**iθ


where e is some number whose value is as yet unknown, which is the solution to the equation:

 e**iπ = -1


As the e() function behaves like an exponential:

e(i*θ1) * e(i*θ2) = e**(i*θ1) * e** (i*θ2) = e**(i*(θ1+θ2)) = e(i(θ1+θ2))


In particular:

 (cos(θ) + i * sin(θ))**n =  e(iθ)**n
= (e**iθ)**n
=  e**inθ
=  e(inθ)
= (cos(nθ) + i * sin(nθ))


Lets try this formula out with n = 2:

  (cos(θ) + i * sin(θ))**2                       = cos(2θ) + i * sin(2θ)


=> (cos(θ)**2 - sin(θ))**2 + 2 * i * cos(θ)sin(θ) = cos(2θ) + i * sin(2θ)

Now, the number i is manifestly not a real number, as no real number is a solution to the equation i*i = -1, yet both the cos() and sin() functions produce real numbered results, they are, after all, just the ratios of the lengths of the sides of triangles. Consequently in the above we can equate the parts of the equations which are separated by being multiplied by i to get two equations:

  cos(2θ) = cos(θ)**2 - sin(θ))**2
sin(2θ) = 2*cos(θ)*sin(θ)


Recall the "Cosine Angle Sum Formula" of:

  cos(θ1+θ2) = cos(θ1)cos(θ2) - sin(θ1)sin(θ2)


Set θ = θ1 = θ2 to get the identical result:

  cos(θ+θ) = cos(2θ) = cos(θ)cos(θ) - sin(θ)sin(θ)


Likewise the Recall the "Sine Angle Sum Formula" of:

  sin(θ1+θ2) = cos(θ1)sin(θ2) + sin(θ1)cos(θ2)


Set θ = θ1 = θ2 to get the identical result:

  sin(θ+θ) = sin(2θ) = cos(θ)sin(θ) + sin(θ)cos(θ) = 2*sin(θ)cos(θ)


Using Cosine and Sine Angle Sum Formulae and equating parts, we can deduce that:

  e(i*θ1) * e(i*θ2))
= (cos(θ1) + i * sin(θ1)) * (cos(θ2) + i * sin(θ2))
= (cos(θ1)*cos(θ2) - sin(θ1)*sin(θ2) + i(cos(θ1)sin(θ2)) + cos(θ2)sin(θ1))
=  cos(θ1 + θ2) + isin(θ1 + θ2)
=  e(i(θ1+θ2))


wherein the e() function demonstrates its exponential nature to perfection.

The ability of i to partition single equations into two orthogonal simultaneous equations makes expressions of the form e(iθ), and hence trigonometery, invaluable in such diverse applications as electronics: simultaneously representing current and voltage in the same equation; and quantum mechanics, where it is necessary to represent position and momentum, or time and energy as pairs of variables partitioned by the uncertainty principle.