# Relationships between exponential function and trigonometric functionsEdit

A similar function is the exponential function which is defined by the statement: the rate of change of is and the value of is 1. Here we only have to apply the rate of change operator once to get back where we started. Explicitly:

tends to as tends to zero.

which can be rewritten replacing by where is any constant number, to get:

tends to as tends to zero.

because is a constant, , performing this substitution, we get:

tends to as tends to zero. tends to as tends to zero.

That is, the rate of change of with is . We can continue this process to find the rate of change of with :

is the limit of: as tends to zero, which is the same as: as tends to zero, which is:

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Performing this 4 time successively yields and comparing with the same action on the function:

If only there was a number , such that , and hence , then we could relate the function to the function . Fortunately, there is a number that will work: the square roots of -1. From here on will denote a square root of -1. is also a solution. We can expect then that is some linear combination of and , perhaps:

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We know that , so:

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Finding the rate of change with :

setting , remembering that ,

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So now we know that and , so , so and .

To summarize what we know so far:

where is in radians and is a square root of -1.

Replacing by gives conversely:

, the same formula, so we must have that .

Given:

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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.