# Constructions for the Golden Ratio

Phi and Trig graphs Here are the graphs of three familiar trigonometric functions: sin x, cos x and tan x in the region of x from 0 to π/2 (radians) = 90°:
The graphs meet at

• the origin, where tan x = sin x
• in the middle, where sin x = cos x i.e. where tan x = 1 or x = 45° = π/4 radians
• at another angle where tan x = cos x

What angle is at the third meeting point?

 tan x = cos x and, since tan x = sin x / cos x , we have: sin x = (cos x)2 =1-(sin x)2 because (sin x)2+(cos x)2=1. or (sin x)2 + sin x = 1

and solving this as a quadratic in sin x, we find
sin x = (–1 + √5)/2 or
sin x = (–1 – √5)/2
The second value is negative and our graph picture is for positive x, so we have our answer:

the third point of intersection is the angle whose sine is Phi – 1 = 0·6180339... = phi

On our graph, we can say that the intersection of the green and blue graphs (cos(x) and tan(x)) is where the red graph (sin(x)) has the value phi [i.e. at the x value of the point where the line y = phi meets the sin(x) curve].

### Is there any significance in the value of tan(x) where tan(x)=cos(x)?

Yes. It is √phi = √0·618033988... = 0·786151377757.. .

Here's how we can prove this.

Take a general right-angled triangle and label one side t and another side 1 so that one angle (call it A) has a tangent of t.
By Pythagoras's Theorem, the hypotenuse is √(1+t2).
So we have:
For all right-angled triangles:
tan A = t
 cos A = 1 √(1 + t2)

 sin A = t √(1 + t2) We now want to find the particular angle A which has tan(A) = cos(A).
From the formulae above we have:

t=1/√(1+t2)

Squaring both sides:

t2 = 1/(1 + t2)

Multiply both sides by 1 + t2

:

t2 ( 1 + t2 ) = 1
t2 + t4 = 1

If we let T stand for t2 then we have a quadratic equation in T which we can easily solve:

t4 + t2 – 1 = 0
T2 + T – 1 = 0
T = ( –1 ± √(1+4) ) / 2

SinceT is t2 it must be positive, so the value of T we want in this case is

T = ( √5 – 1 ) / 2 = phi Since T is t2 then t is √phi.
Since 1 + Phi = Phi2 then the hypotenuse √(1+T2) = Phi as shown in the triangle here.
From the 1, √Phi, Phi triangle, we see that

tan(A)=cos(A)
tan(B)=1/sin(B) or tan(B)=cosec(B)

What is the angle whose tangent is the same as it cosine?

In the triangle here, it is angle A:
A = 38.1727076° = 30° 10' 21.74745" = 0.1060352989.. of a whole turn = 0.666239432.. radians.
The other angle, B has its tangent equal to its cosecant (the reciprocal of the sine):
B = 90° – A = 51.8272923..° = 51° 49' 38.2525417.." = 0.143964701.. of a whole turn = 0.90455689.. radians.

Notice how, when we apply Pythagoras' Theorem to the triangle shown here with sides 1, Phi and root Phi, we have

Phi2 = 1 + (√Phi)2 = 1 + Phi

which is one of our classic definitions of (the positive number) Phi.....

The next section looks at the other trigonometrical relationships in a triangle and shows that, where they are equal, each involves the numbers Phi and phi.

#### Notation for inverse functions

A common mathematical notation for the-angle-whose-sin-is 0.5 is arcsin(0.5) although you will sometimes see this written as asin(0.5).
We can prefix arc (or a) to any trigonometrical function (cos, cot, tan, etc.) to make it into its "inverse" function which, given the trig's value, returns the angle itself.
Each of these inverse functions is applied to a number and returns an angle.
e.g. if sin(90°)=0 then 0 = arcsin(90°).

What is arccos(0.5)?
The angle whose cosine is 0.5 is 60°.
But cos(120°)=0.5 as does cos(240°) and cos(300°) and we can add 360° to any of these angles to find some more values!
The answer is arccos(0.5) = 60° or 120° or 240° or 300° or ...
With all the inverse trig. functions you must carefully select the answer or answers that are appropriate to the problem you are solving.

"The angle whose tangent is the same as its cosine" can be written mathematically in several ways:

tan(A)=cos(A) is the same as arctan( cos(A) ) = A

Can you see that it can also be written as arccos( tan(A) )= A?

### More trig ratios and Phi (sec, csc, cot)

In a right-angled triangle if we focus on one angle (A), we can call the two sides round the right-angle the Opposite and the Adjacent sides and the longest side is the Hypotenuse, or Adj, Opp and Hyp for short. You might wonder why we give a name to the ratio Adj/Opp (the tangent) of angle A but not to Opp/Adj. The same applies to the other two pairs of sides: we call Opp/Hyp the sine of A but what about Hyp/Opp? Similarly Adj/Hyp is cosine of A but what about Hyp/Adj?

In fact they do have names:

• the inverse ratio to the tangent is the cotangent or cot i.e. Adj/Opp; cot(x)=1/tan(x)
• the inverse ratio to the cosine is the secant or sec i.e. Hyp/Adj; sec(x)=1/cos(x)
• the inverse ratio to the sine is the cosecant or cosec or sometimes csc i.e. Hyp/Opp; csc(x) = 1/sin(x)

You'll notice that these six names divide into two groups:

• secant, sine, tangent
• cosecant, cosine, cotangent

and show another way of choosing one representative for each of the 3 pairs of ratios (x/y and y/x where x and y are one of the three sides).

Here is a graph of the six functions where the angle is measured in radians: This extended set of graphs reveals two more intersections involving Phi, phi and their square roots: e.g.

• if A is the angle where cos(A) = tan(A) then sin(A) = phi and cosec(A) = Phi;
The value of A is A = 38.172..° = 0.666239... radians;
• if B is the angle where sin(B) = cot(B) then cos(B) = phi and sec(B) = Phi;
The value of B is 51.827..° = 0.904556... radians.

Notice that A and B sum to 90° - as we would expect of any two angles where the sine of one is the cosine of the other. Some Results in Trigonometry L Raphael, Fibonacci Quarterly vol 8 (1970), pages 371-392.

#### Things to do 1. Above we solved cos(x)=tan(x) using the 1,t,√(1+t2) triangle. Use the same triangle and adapt the method to find the value of sin(x) for which sin(x)=cot(x).
If you use the formulae above then remember that you will find t, the tangent of the angle for which sin(x)=cot(x). Since we want the cotangent, just take the reciprocal of t to solve sin(x)=cot(x)=1/t.
2. On the previous page we saw two ways to find Phi on your calculator using the 1/x button, square-root button and just adding 1. Here's how we can do the same thing to find √phi and √Phi.
From tan(x)=cos(x) we found t=phi is a solution to t=1/√(1+t2)
1. Enter any number to start the process
1. Square it
3. Take the square-root
4. Take the reciprocal (the 1/x button)
5. Write down the number now displayed
2. Repeat the previous step as often as you like.
Each time, given t, we compute 1/√(1+t2) and keep repeating this transformation.
Eventually, the number we write down does not change. It is √phi = 0.7861513777... . In fact, no matter how big or small is your starting value, you'll get √phi to 4 or 5 dps after only a few iterations. Try it!
If you want √Phi, just use the 1/x button on your final answer. Some Results in Trigonometry, Brother L Raphael, The Fibonacci Quarterly vol 8 (1970) pages 371 and 392.

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