Construction of Right Triangles

Right triangles are easily constructed. Recall that a diameter is a straight line which starts at one point on the circle and goes through the center to the other side. Using this property of a circle:

Construct a diameter of a circle. Call the points where the diameter touches the circle a and b.
Choose a distinct point c on the circle (any point that is neither a nor b).
Construct line segments connecting points a, b and c.

Provided the above three directions are followed, the resulting triangle Δabc will be a right triangle. Note that the right angle will be formed on the circle unless intentionally designated as the single instance at which it exists at the center point (e.g. when the circle is divided into equal quadrants by two perpendicular diameters).

This right triangle can be further divided into two isosceles triangles by adding a line segment from c to the center of the circle.

To simplify the following discussion, we specify that the circle has diameter 1 and is oriented such that the diameter drawn above runs left to right. We shall denote the angle of the triangle at the right θ and that at the left φ. The sides of the right triangle will be labeled, starting on the right, a, b, c, so that a is the rightmost side, b the leftmost, and c the diameter. We know from earlier that side c is opposite the right angle and is called the hypotenuse. Side a is opposite angle φ, while side b is opposite angle θ. We reiterate that the diameter, now called c, shall be assumed to have length 1 except where otherwise noted.

By constructing a right angle to the diameter at one of the points where it crosses the circle and then using the method outlined earlier for producing binary fractions of the right angle, we can construct one of the angles, say θ, as an angle of known measure between 0 and π / 2. The measure of the other angle φ is then π - π/2 - θ = π/2 - θ, the complement of angle θ. Likewise, we can bisect the diameter of the circle to produce lengths which are binary fractions of the length of the diameter. Using a compass, a binary fractional length of the diameter can be used to construct side a (or b) having a known size (with regard to the diameter c) from which side b can be constructed.

Using Right Triangles to find Unknown Sides

Finding unknown sides from two other sides

From the Theorem of Pythagoras, we know that:

a^{2} + b^{2} = c^{2} .

Recall that c (the diameter in the above example) has been defined as having a length of one, therefore:

b^{2} = 1 - a^{2} .

permits us to calculate the length of b squared. It may happen that b squared is a fraction such as 1/4 for which a square root can be explicitly found, in this case as 1/2, alternatively, we could use Newton's Method to find an approximate value for b.

Finding unknown sides from a side and a (non-right) angle

As any triangle could be compared to our basic triangle (formed from a circle with a diameter of one), a table enumerating the relationships between angles and side lengths would be very useful to understanding the properties of any triangle. However, such a table would be unwieldy in practice and it is often not necessary to know the exact value.

Of course, given an angle $θ$ , we could construct a right angled triangle using ruler and compass that had $θ$ as one of its angles, we could then measure the length of the side that corresponds to a to evaluate the $\cos$ function. Such measurements would necessarily be in-exact; it would be a problem in physics to see how accurately such measurements can be made; using trigonometry we can make precise predictions with which the results of these physical measurements can be compared.

A notation would allow us to recognize that we're using a certain relationship without having to know the exact numerical values. We call such a notation a function. Basic trigonometric functions are simply stand-ins for the relationship between angles and sides of a triangle.

One such function, which allows us to know the relationship between any value of $θ$ and the corresponding value of $a$ is called the cosine $\cos$ . This universal relationship is represented as: $\frac{a}{c} = \cos θ$ which would save us the work of constructing angles and lengths and making difficult deductions from them.

This means that if you know the cosine of an angle, you also know the relationship between the lengths of the sides. The actual size of the triangle can be bigger or smaller, but the mathematical relationship represented by the cosine does not change so long as the size of the angle remains constant.

Cosine example

Some explicit values for the $\cos$ function are known. For $θ = 0$ , sides $a$ and $c$ are coincidental: $a = c = \frac{a}{c} = 1$ , so $\cos 0 = 1$ . For $θ = \frac{π}{2}$ , sides $b$ and $c$ are coincidental and of length 1, and side $a$ is of zero length, consequently $a = 0$ , $c = 1$ . $\frac{a}{c} = 0$ and $\cos \frac{π}{2} = 0$ .

The simplest right angle triangle we can draw is the isoceles right angled triangle, it has a pair of angles of size $\frac{π}{4}$ radians, and if its hypotenuse is considered to be of length one, then the sides $a$ and $b$ are of length $\sqrt{\frac{1}{2}}$ as can be verified by the theorem of Pythagoras. If the side $a$ is chosen to be the same length as the radius of the circle containing the right angled triangle, then the right hand isoceles triangle obtained by splitting the right angled triangle from the circumference of the circle to its center is an equilateral triangle, so $θ$ must be $\frac{π}{3}$ , and $ϕ$ must be $\frac{π}{6}$ and $b^{2}$ must be $1 - \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$ .

Properties of the cosine and sine functions

Period

Ancient observation of the circle (and its definition of perfect roundness) lead to the finding that circles increase in size by the value of 2π. Therefore, a perfect circle is maintained, along with the relationships formed by triangles within, by adding 2π to any angle θ. This is called the period, --the size of the angle or the time period over which the relationships begin to repeat (correlating the two is complex, and allows us to talk about wave theory).

Using functions, we can represent this fact in terms of the cosine function by stating that : cos(θ+2πn) = cos(θ).

Knowing the period of the sine and cosine functions (and by derivation, that of other functions) is useful because it means that we can substitute one angle for another when we know that the period is the same. This helps in calculations, such as when there is the need to add or subtract angles.

Half Angle and Double Angle Formulas

We can derive a formula for cos(θ/2) in terms of cos(θ) which allows us to find the value of the cos() function for many more angles. To derive the formula, draw an isoceles triangle, draw a circle through its corners, connect the center of the circle with radii to each corner of the isoceles triangle, extend the radius through the apex of the isoceles triangle into a diameter of the circle and connect the point where the diameter crosses the other side of the circle with lines to the other corners of the isoceles triangle.

Therefore:

$\cos (θ / 2) = \sqrt{\frac{1 + \cos θ}{2}}$

which gives a method of calculating the cos() of half an angle in terms of the cos() of the original angle. For this reason * is called the "Cosine Half Angle Formula".

The half angle formula can be applied to split the newly discovered angle which in turn can be split again ad infinitum. Of course, each new split involves finding the square root of a term with a square root, so this cannot be recommended as an effective procedure for computing values of the cos() function.

Equation * can be inverted to find cos(θ) in terms of cos(θ)/2:

$\begin{matrix} \cos (θ / 2) = \sqrt{\frac{1 + \cos (θ)}{2}} \\ \Rightarrow & \cos^{2} (θ / 2) = \frac{1 + \cos (θ)}{2} \\ \Rightarrow & \cos^{2} (θ / 2) - \frac{1}{2} = \frac{\cos (θ)}{2} \\ \Rightarrow & 2 \cos^{2} (θ / 2) - 1 = \cos (θ) \end{matrix}$

substituting $δ = \frac{θ}{2}$ gives:

$2 \cos^{2} (δ) - 1 = \cos (2 δ)$

that is a formula for the cos() of double an angle in terms of the cos() of the original angle, and is called the "cosine double angle sum formula".

Angle Addition and Subtraction formulas

To find a formula for $\cos (θ_{1} + θ_{2})$ in terms of $\cos θ_{1}$ and $\cos θ_{2}$ : construct two different right angle triangles each drawn with side $c$ having the same length of one, but with $θ_{1} \neq θ_{2}$ , and therefore angle $ψ_{1} \neq ψ_{2}$ . Scale up triangle two so that side $a_{2}$ is the same length as side $c_{1}$ . Place the triangles so that side $c_{1}$ is coincidental with side $a_{2}$ , and the angles $θ_{1}$ and $θ_{2}$ are juxtaposed to form angle $θ_{3} = θ_{1} + θ_{2}$ at the origin. The circumference of the circle within which triangle two is imbedded (circle 2) crosses side $a_{1}$ at point $g$ , allowing a third right angle to be drawn from angle $φ_{2}$ to point $g$ . Now reset the scale of the entire figure so that side $c_{2}$ is considered to be of length 1. Side $a_{2}$ coincidental with side $c_{1}$ will then be of length $\cos θ_{1}$ , and so side $a_{1}$ will be of length $\cos θ_{1} \cdot \cos θ_{2}$ in which length lies point $g$ . Draw a line parallel to line $a_{1}$ through the right angle of triangle two to produce a fourth right angle triangle, this one imbedded in triangle two. Triangle 4 is a scaled copy of triangle 1, because:

 (1) it is right angled, and 
 (2)  $θ_{4} + (π - \frac{π}{2} - θ_{1} - θ_{2}) = φ_{2} = π - \frac{π}{2} - θ_{2} \Rightarrow θ_{4} = θ_{1}$ .

The length of side b4 is cos(φ2)cos(φ4) = cos(φ2)cos(φ1) as θ4 = θ1 Thus point g is located at length:

 cos(θ1+θ2) = cos(θ1)cos(θ2) - cos(φ1)cos(φ2), where θ1+φ1 = θ2+φ2 = π/2

giving us the "Cosine Angle Sum Formula".

Proof that angle sum formula and double angle formula are consistent

We can apply this formula immediately to sum two equal angles:

   cos(2θ) = cos(θ+θ) = cos(θ)cos(θ) - cos(φ)cos(φ) where θ+φ = π/2
           = cos(θ)**2 - cos(φ)**2           (I)

From the theorem of Pythagoras we know that:

   a**2 + b**2 = c**2

in this case:

   cos(θ)**2 + cos(φ)**2  = 1**2                  where θ+φ = π/2
 =>            cos(φ)**2  = 1 - cos(θ)**2

Substituting into (I) gives:

   cos(2θ)   = cos(θ)**2 -      cos(φ)**2         where θ+φ = π/2 
             = cos(θ)**2 - (1 - cos(θ)**2)
             = 2 * cos(θ)**2 - 1

which is identical to the "Cosine Double Angle Sum Formula":

   2 * cos(δ)**2 - 1 = cos(2δ)

Derivative of the Cosine Function

We are now in a position to evaluate the expression:

 cos(θ+δθ)/ δθ = (cos(θ)cos(δθ) - cos(δθ)cos(δφ)) /δθ   where θ+φ = δθ+δφ = π/2.

If we let δθ tend to zero we get an increasingly accurate expression for the rate at which cos(θ) varies with θ. If we let δθ tend to zero, cos(δθ) will tend to one, δφ will tend to π/2, cos(δφ) will tend to zero, and cos(δφ)/δθ will tend to one: allowing us to write in the limit:

 (cos(θ+δθ) - cos(θ))/δθ = (cos(θ)cos(δθ)-cos(φ)cos(δφ)-cos(θ)) /δθ   where θ+φ = δθ+δφ = π/2.
                         = (cos(θ) * 1   -cos(φ) * δθ  -cos(θ)) /δθ 
                         = -cos(φ) * δθ /δθ 
                         = -cos(φ)                                    where θ+φ = π/2.
                         = -cos(π/2 - θ)

The function cos(π/2 - θ) occurs with such prevalence that it has been given a special name:

 sin(θ) = cos(π/2 - θ) 
        = cos(φ) where θ+φ = π/2.

"sin" is pronounced "sine".

The above result can now be stated more succinctly:

 (cos(θ+δθ) - cos(θ))/δθ = -sin(θ)

often further abbreviated to:

 d cos(θ)/δθ = -sin(θ)

or in words: the rate of chage of cos(θ) with θ is -sin(θ).

Pythagorean identity

Armed with this definition of the sin() function, we can restate the Theorem of Pythagoras for a right angled triangle with side c of length one, from:

    cos(θ)**2 + cos(φ)**2 = 1**2  where θ+φ = π/2

to:

    cos(θ)**2 + sin(θ)**2 = 1.

We can also restate the "Cosine Angle Sum Formula" from:

 cos(θ1+θ2) = cos(θ1)cos(θ2) - cos(φ1)cos(φ2), where θ1+φ1 = θ2+φ2 = π/2

to:

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

Sine Formulas

The price we have to pay for the notational convenince of this new function sin() is that we now have to answer questions like: Is there a "Sine Angle Sum Formula". Such questions can always be answered by taking the cos() form and selectively replacing cos(θ)**2 by 1 - sin(θ)**2 and then using algebra to simplify the resulting equation. Applying this technique to the "Cosine Angle Sum Formula" produces:

   cos(θ1+θ2)        =      cos(θ1)cos(θ2) - sin(θ1)sin(θ2)
=> cos(θ1+θ2)**2     =     (cos(θ1)cos(θ2) - sin(θ1)sin(θ2))**2
=> 1 - cos(θ1+θ2)**2 = 1 - (cos(θ1)cos(θ2) - sin(θ1)sin(θ2))**2
=> sin(θ1+θ2)**2     = 1 - (cos(θ1)**2*cos(θ2)**2 + sin(θ1)**2*sin(θ2)**2 - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2))   -- Pythagoras on left, multiply out right hand side

                     = 1 - (cos(θ1)**2*(1-sin(θ2)**2) + sin(θ1)**2*(1-cos(θ2)**2) - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Carefully selected Pythagoras again on the left hand side

                     = 1 - (cos(θ1)**2-cos(θ1)**2*sin(θ2)**2 + sin(θ1)**2-sin(θ1)**2*cos(θ2)**2 - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Multiplied out

                     = 1 - (1         -cos(θ1)**2*sin(θ2)**2             -sin(θ1)**2*cos(θ2)**2 - 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Carefully selected Pythagoras

                     =                 cos(θ1)**2*sin(θ2)**2              sin(θ1)**2*cos(θ2)**2 + 2*cos(θ1)cos(θ2)sin(θ1)sin(θ2)) -- Algebraic simplification

                     =                (cos(θ1)sin(θ2)                    +sin(θ1)cos(θ2))**2

taking the square root of both sides produces the the "Sine Angle Sum Formula"

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

We can use a similar technique to find the "Sine Half Angle Formula" from the "Cosine Half Angle Formula":

cos(θ/2) = squareRoot(1/2 + cos(θ)/2).

We know that 1 - cos(θ/2)**2 = sin(θ/2) **2, so squaring both sides of the "Cosine Half Angle Formula" and subtracting from one:

 => (1 - cos(θ/2)**2) =       1 - (1/2 + cos(θ)/2)      
 =>      sin(θ/2)**2  =            1/2 - cos(θ)/2      
 =>      sin(θ/2)     = squareRoot(1/2 - cos(θ)/2)

So far so good, but we still have a cos(θ/2) to get rid of. Use Pythagoras again to get the "Sine Half Angle Formula":

         sin(θ/2)     = squareRoot(1/2 - squareRoot(1 - sin(θ)**2)/2)

or perhaps a little more legibly as:

         sin(θ/2)     = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - sin(θ)**2))

On the Use of Computers for Algebraic Manipulation

The algebra to perform such manipulations tends to be horrendous: the horrors can be reduced by using automated symbolic algebraic manipulation software to confirm that each line of the derivation has been carried out correctly. However, such software is unable, in of itself, to perform the set of deductions above without human intervention: you will note that at various points in the above sequence, the Theorem of Pythagoras was cunningly applied for reasons that only became obvious towards the end of the derivation.

Alternatively, one can draw a plausible geometric diagram, make careful deductions about unknown sides and angles from known sides and angles. Again, software can check that each inference is consistent with the known facts, however, such software cannot decide which diagram to draw in the first place, nor which are the significant inferences to be made. You will no doubt have experienced the alarming feeling when manipulating such equations or diagrams that the numbers of terms in the equations, or the numbers of lines, sides and circles on the diagram, are doubling after each possible inference, multiplying exponentially beyond your control, with no solution in sight. Undirected software rapidly looses itself in this exponential maze despite the protestations of the salaried scientists held captive by Microsoft at Cambridge University, who protest, I think too much, that Alan Turing and Kurt Godell were wrong, and that "Machines can do Mathematics" if only they are given enouigh time to program (that is direct) them, to do so. Fortunately, the most that machines can do, is: assist humans with the tedious but necessary task of verifying each logical deduction. Machines cannot see the beauty of geometry, nor find directions to solve equations.

As is possible to define many other ratios of sides in an right angled triangle, the ratio of side b to side a for example. There will surely be relationships between these newly defined auxillary trigonometric functions - it would be very odd if there were not. However, all such subsidary functions can be dealt with by the properties of the cos() function alone; any newly defined functions must earn their keep by reducing complexity, not increasing it: their existence often creates more definitions to learn without generating any significantly new knowledge. Mathematics should be derivable from as few ideas as possible, that is from quality without quantity. Those of you, other than the physics students liberated by Feynmann in Brazil, who are facing exams produced at the behest of tyrannical governments will be reduced to the tedium of learning the many intricate relationships between these auxillary functions, so that a bureaucrat with a fetish for metrics can kowtow to public opinion: please keep in mind that quantity and quality are not necessarily the same in mathematics; that increasing the number of auxiliary functions and their interrelationships merely aggravates the number of possibilities in each step of the solution space, without offering you any useful guidance as to which would be the most to use.

Sine derivative

Relationships between exponential function and trigonometric functions

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

 (e(θ+δθ)-e(θ)) / δθ tends to e(θ) as δθ tends to zero.

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

 (e(iθ+δiθ)-e(iθ)) / δiθ tends to e(iθ) as δθ tends to zero.

because i is a constant, δiθ = iδθ, performing this subsitution, we get:

    (e(iθ+iδθ)-e(iθ))  / iδθ tends to e(iθ) as δθ tends to zero.
=>  (e(i(θ+δθ))-e(iθ)) / iδθ tends to e(iθ) as δθ tends to zero.
=>  d e(iθ) / dθ  = i*e(iθ)

That is, the rate of change of e(iθ) with θ is i*e(iθ). We can continue this process to find the rate of change of i*e(iθ) with θ:

  d i*e(iθ) / dθ                                    is the limit of: 
  (i* e(i(θ+δθ))-i*e(iθ)) / δθ as δθ tends to zero, which is the same as:
   i*(e(i(θ+δθ))-  e(iθ)) / δθ as δθ tends to zero, which is:

   i* d e(iθ) / dθ = i*i*e(iθ).

Performing this 4 time sucessively yields and comparing with the same action on the cos() function:

   d       e(iθ) / dθ  =       i*e(iθ)               d   cos(θ) / dθ  = - sin(θ)
   d     i*e(iθ) / dθ  =     i*i*e(iθ)               d - sin(θ) / dθ  = - cos(θ)
   d   i*i*e(iθ) / dθ  =   i*i*i*e(iθ)               d - cos(θ) / dθ  =   sin(θ)
   d i*i*i*e(iθ) / dθ  = i*i*i*i*e(iθ)               d   sin(θ) / dθ  =   cos(θ)

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

 cos(θ) = A*e(iθ)+B*e(-iθ)

We know that cos(0) = 1, so:

 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 θ:

=> d cos(θ) / dθ = d (A*e(iθ)) / dθ +d  (B*e(-iθ))/dθ 
=> - sin(θ)      =  i*A*e(iθ)       + -i*B*e(-iθ) = - cos(π/2-θ)

setting θ = 0, remembering that cos(π/2) = 0

=> 0 = iA - iB = i(A-B) => A = B

So now we know that A+B = 1 and A = B, so A+A = 1, so A = 1/2 and B = 1/2.

To summarize what we know so far:

cos(θ) = e(iθ)/2 + e(-iθ)/2   where θ is in radians and i is a square root of -1.

Replacing θ by -θ gives conversly:

cos(-θ) = e(-iθ)/2 + e(iθ)/2, the same formula, so we must have that cos(θ) =  cos(-θ).

Given:

cos(θ) = e(iθ)/2 + e(-iθ)/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;

Similarily, 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 algrebraic 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 exponeniation, 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.

Taylor series approximations for the trig. functions

Having learnt a good deal about the properties of the cos() function, it might be helpful to know how to calculate cos(θ) for a given θ. We know that cos() can be defined in terms of the e() function which has the remarkable property that d e(θ) / dθ = e(θ), and e(0) = 1. Perhaps it is possible to represent e() by an infinite polynomial:

e(θ) = a0 + a1*θ + a2*θ**2 + a3*θ**3 + ...  an*θ**n +  ....

in which case evaluating at θ = 0 yields a0 = 1. To find the rate of change with θ of the n'th term:

d an*θ**n / dθ = (an*(θ+δθ)**n - an*(θ)**n)           / δθ as δθ tends to zero
               = an *((θ+δθ)**n - θ**n)               / δθ as δθ tends to zero
               = an *((θ**n+n*δθ*θ**(n-1)+... - θ**n) / δθ as δθ tends to zero
               = an *(n*δθ*θ**(n-1)+...)              / δθ as δθ tends to zero
               = an * n*θ*(n-1)+...

where the ... represent terms multiplied by at least δθ**2, and which therefore tend to zero as δθ tends to zero.

Finding the rate of change with respect to θ of the polynomial representing e(θ) produces:

e(θ) = a1*θ + 2 * a2*θ**1 + 3 * a3*θ**2 + ...  n * an*θ**(n-1) +  ....

Again evaluating at θ = 0, we get a1 = 1

Repeating the whole process, we get:

e(θ) = 2 * a2 + 3 * 2* a3*θ + ...  n * (n-1) * an*θ**(n-2) +  ....

and evaluating at θ = 0, we get a2 = 1/2, and in general, an = 1/n!, wher n! means n*(n-1)*(n-2)*...3*2*1.

Putting these results together, we get:

e(θ) = 1 + θ + θ**2/(2*1) + θ**3/(3*2*1)2 + ... θ**n / n! +  ....

Recalling that e(θ) = e**θ, and thus that e(1) = e**1 = e, we find e to be:

e = 1+1+1/2+1/6+1/24+.... 1/n!+ ...

which is approximately equal to 2.71828183

To calculate cos(θ), we would apply the formula:

   2*cos(θ) = e(iθ)+e(-iθ)
            = 1 + iθ - θ**2/2 - iθ**3/6 + θ**4/24 + ...   
            + 1 +-iθ - θ**2/2 --iθ**3/6 + θ**4/24 + ...
            = 2      -2θ**2/2           +2θ**4/24 + ...

 => cos(θ)  = 1 - θ**2/2 + θ**4/24 + ... (-1)**n*θ**(2n)/((2n)!)+ ...

in which the odd terms cancel out and the even terms double and alternate in sign. Substituting -θ for θ does not change the result as θ occurs with even powers, hence the earlier phrase that cos() is an even function.

Using the Taylor series to calculate π

Having calculated, e and cos(), lets see if we can calculate π. Draw a circle with radius of length one, and draw diameters to divide the circle up into n segments of equal area, and hence equal angles θ; connect the ends of each diameter to its neighbors to get a regular polygon of n equal length sides. Each segment of the polygon is an isoceles triangle imbedded within a segment of the circle. The length of the sides og the polygon is less than the distance along the circumference of the circle between each corner of the polygon, the circumference length is exactly θ because we are using radians to measure angles within a circle of radius of length one. We can bisect each side of the polygon to construct a regular polygon with 2*n equal sides,the angle between each diameter is now θ/2 - lets us call the angle between diameters the interior angle. As the interior angle gets smaller and smaller, the lengths measured between two adjacent corners of the polygon, measured first along the side of the polygon, the choord length, and secondly along the circumference of the circle, the arc length, agree more and more closely. We can get this agreement as close as we might like by dividing the interior angle often enough, although to get eact agreement we would have to divide the interior angle an infinte number of times, clearly impossible in the real world of physics, but within the realm of mathematics it is possible to contemplate such an infinite process.

We have already defined the length of the circumference of a circle of diameter of length one as having the value π, so now all we need to do is find the total length of the sides of the polygons we are using to approximate π and then extrapolate as the number of sides approaches infinity to get an estimate for the value of π.

Let us start the process off using a polygon of 4 sides, a square, which has an interior angle of a right angle, that is, π/2. Draw a square, then draw its diagonals, and a circle through the corners of the square, to get 4 chords and 4 arcs. Scale the drawing so that the radius of the circle drawn is exactly 1/2. Each chord is the base of isoceles triangle which can be bisected into two identical right angled triangles. The hypotenuse or side c of of each of these right angle triangles is a radius, we have scaled the drawing above so that each such radius will be of length 1/2. The angle θ of each such right angle triangle is 1/2*π/2 = π/4. The length of side b in each such right angle triangle has length 1/2*sin(π/4) = 1/2*sin(2π/8). If we bisect each interior angle of the square we get a total of 8 of these right angled triangles, hence our first approximation to the length of the circumference of a circle of diameter of length one which is defined to be π, is 8 * the length of side b:

    π = 8 * 1/2 * sin(2*π/8)
=> 2π = n * sin(2*π/n) where n = 8.

If we extend each side a of each of the each such right angle triangles to meet the circumference of the circle at points, and then draw lines from each of these points to the points neighboring points where the diagonals of the square cross the circumference we will construct the octogon - a polygon with 8 equal sides - twice the number of the square, with interior angles of π/4, half that of the square. The argument used above to calculate the lengths of the sides of the square can be applied to the octogon. When this argument was first applied to the square it seems unnecessarily complicated: it would have been easier to notice that diagonals of the square conveniently meet at right angles allowing us to apply Pythagoras immediately to calculate the length of the side of a square drawn in a circle of radius of length 1/2 as squareRoot(1/2**2+1/2**2) = squareRoot(1/4+1/4) = squareRoot(1/2). While this answer is perfectly correct for a square, it does not work for any other polygon as their diagonals do not meet at such a convenient angle, while the formula:

=> 2π = n * sin(2*π/n)

works for any value of n.

All we need now is the value of the sin() function for polygions with many sides, that is with very small interior angles, or more specifically, with n large. We do know the sin() function for one angle:

sin(π/2) = sin(2π/4) = 1

The "Sine Half Angle Formula" lets us calculate the value of the sin() finction for half angle of known angles:

    sin(θ/2)      = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - sin(θ)**2))  
  
 => sin(2π/8)     = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - 1**2))      
                  = squareRoot(1/2)

 => sin(2π/16)    = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - squareRoot(1/2)**2))      
                  = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - 1/2))      
                  = squareRoot(1/2 - 1/2*squareRoot(1/2))

 => sin(2π/32)    = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - (squareRoot(1/2 - 1/2*squareRoot(1/2)))**2))  
                  = squareRoot(1/2)*squareRoot(1 - squareRoot(1 - 1/2 + 1/2*squareRoot(1/2)))  
                  = squareRoot(1/2)*squareRoot(1 - squareRoot(1/2 + 1/2*squareRoot(1/2)))  
                  = squareRoot(1/2 - 1/2*squareRoot(1/2 + 1/2*squareRoot(1/2)))

Its easy to guess from this pattern that:

 => sin(2π/64)    = squareRoot(1/2 - 1/2*squareRoot(1/2 + 1/2*squareRoot(1/2 + 1/2*squareRoot(1/2))))

and so on. This formula does give an approximate value for π, an 8 digit calculator produced a value of 3.13 which is within 1% of the true value. However, this formula cannot be ergarded as an efficient way of calculating π accuartely to many digits: the constant need to take square roots insures a continual loss of precision.

We have now built up some basic trigonometric results.

Trigonometric defintitions

We have defined the sine, cosine, and tangent functions using the unit circle. Now we can apply them to a right triangle.

This triangle has sides a and b. The angle between them, C, is a right angle. The third side, c, is the hypotenuse. Side a is opposite angle A, and side b is adjacent to angle A.

Applying the definitions of the functions, we arrive at these useful formulas:

   sin(A) = a / c  or opposite side over hypotenuse

   cos(A) = b / c  or adjacent side over hypotenuse

   tan(A) = a / b  or opposite side over adjacent side

This can be memorized using the mnenomic 'SOHCAHTOA' (sin = opposite over hypothenuse, cosine = adjacent over hypotenuse, tangent = opposite over adjacent).

Exercises: (Draw a diagram!)

1. A right triangle has side a = 3, b = 4, and c = 5. Calculate the following:

sin(A), cos(A), tan(A)

2. A different right triangle has side c = 6 and sin(a) = 0.5 . Calculate side a.

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Trigonometry/Right angle trigonometry

Contents

Construction of Right Triangles

Using Right Triangles to find Unknown Sides

Finding unknown sides from two other sides

Finding unknown sides from a side and a (non-right) angle

Cosine example

Properties of the cosine and sine functions

Period

Half Angle and Double Angle Formulas

Angle Addition and Subtraction formulas

Proof that angle sum formula and double angle formula are consistent

Derivative of the Cosine Function

Pythagorean identity

Sine Formulas

On the Use of Computers for Algebraic Manipulation

Sine derivative

Relationships between exponential function and trigonometric functions

Taylor series approximations for the trig. functions

Using the Taylor series to calculate π

Trigonometric defintitions

Navigation menu

Trigonometry/Right angle trigonometry

Construction of Right Triangles

Using Right Triangles to find Unknown Sides

Finding unknown sides from two other sides

Finding unknown sides from a side and a (non-right) angle

Cosine example

Properties of the cosine and sine functions

Period

Half Angle and Double Angle Formulas

Angle Addition and Subtraction formulas

Proof that angle sum formula and double angle formula are consistent

Derivative of the Cosine Function

Pythagorean identity

Sine Formulas

On the Use of Computers for Algebraic Manipulation

Sine derivative

Relationships between exponential function and trigonometric functions

Taylor series approximations for the trig. functions

Using the Taylor series to calculate π

Trigonometric defintitions

Navigation menu

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