# Algebraic proofs of trigonometric identities

1. May 15, 2012

### KeithLucas

Hello all,

I was wondering if someone has ever found a purely algebraic proof for the addition/subtraction theorems of trigonometry, mainly sin(a+b)=sin(a)cos(b)+sin(b)cos(a). Given a right triangle:

Let x be one of the perpendicular legs and let the other leg be composed of two parts, y1 and y2. Let the line that separates angle a from angle b also be the line that separates y1 from y2 when it intersects with the perpendicular leg. Let a line segment that is perpendicular to the hypotenuse be drawn to intersect with the leg composed of y1 and y2. Call the parts of the hypotenuse on each side of the line z1 and z2 respectively.

Eq1) (z1+z2)^2 = x^2 + (y1+y2)^2
Eq2) x^2 + y1^2 = z1^2 + y2^2 - z2^2

Prove: (y1+y2)/(z1+z2)=(y1z1)/(x^2 + y1^2) + (x(y2^2 - z2^2)^0/5)/(x^2 + y1^2)

1) Any thoughts?
2) Is using latex preferable?

2. May 15, 2012

### coolul007

3. May 16, 2012

### Skrew

It depends upon how you define sine/cosine.

If you define them using infinite series or differential equations it is impossible to give an "algebraic proof" as you have described. They are no longer the values of a picture or geometric objects, the picture and geometric objects is of the values. If that makes sense.

4. May 16, 2012

### coolul007

The Law of Cosines is the algebraic proof...

5. May 16, 2012

### Office_Shredder

Staff Emeritus
Euler's formula letrs you prove this using zero geometry

6. May 16, 2012

### SteveL27

My favorite is the proof based on the Euler relation for the complex exponential,

$e^{it} = cos(t) + i * sin(t)$

We have

$cos(t_1+t_2) + i * sin(t_1 + t_2)$

$= e^{i*(t1 + t2)}$

$= e^{i*t_1} * e^{i*t_2}$

$= (cos(t_1) + i*sin(t_1)) * (cos(t_2) + i * sin(t_2))$

$= cos(t_1) * cos(t_2) - sin(t_1) * sin(t_2) + i (cos(t_2) * sin(t_1) + cos(t_1) * sin(t_2) )$

and then equating the real and imaginary parts of the first and last lines gives the addition formulas for cos and sin.

I suppose that's not "algebraic" in the sense that it involves the complex exponential function, but it's still a nice way to derive the formulas.

You beat me to this while I was hacking the LaTeX :-)

7. May 16, 2012

### KeithLucas

The complex exponential function use is fine, but the theorem doesn't fit what I meant by Algebra. By Algebra, I mean, without using limits or geometry other than the pythagorean identities I gave above. Euler's formula is only derived from using limits and differentiation. Maybe I need to draw a picture. How would I post a picture to the forum?

8. May 16, 2012

### jgutierrez218

Law of Cosines is {x|-1 ≤ y ≤ +1} ????

9. May 16, 2012

### coolul007

c^2 = a^2 + b^2 -2abCos(C)

10. May 16, 2012

### mesa

You are going to have to be a little more specific.

Where is y1 and y2 respective to the side closer to the right angle?
Which side is x and y1 y2? Is it the larger x or y1 y1? Or are they the same length?
Where is z1 and z2 with respect to y1 and y2?
Is z1 z2 the hypotenuses created from the line perpendicular to the origonal hypotenuse?
The line perpendicular to the large hypotenuse, I'm assuming, should intersect the other perpendicular side (y1, y2) where the two line segments start, is that correct?
Finally which side of the angle are a and b on?

Or you could post a pic! ;)

11. May 25, 2012

### KeithLucas

I've uploaded a picture that's of admittedly very poor quality, but I think it does the job. I'm trying to use the algebraic relationships to demonstrate the trigonometric addition/subtraction theorem.

Click on the thumbnail to view the picture.

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12. May 25, 2012

### mesa

Yikes, you may want to re-take that picture :)

13. May 25, 2012

### Cardinal

I think the following is what you're looking for...

so x1=cosβ; y1=sinβ; x2=cosα; y2=sinα

distance between M1 and M2 is:

d2=(x2-x1)2+(y2-y1)2=(cosα-cosβ)2+(sinα-sinβ)2=cos2α-2cosαcosβ+cos2β+sin2α-2sinαsinβ+sin2β=2-2cosαcosβ-2sinαsinβ

Now let's make a coordinate system Ox'y' rotated ccw at angle β with respect to the original one, Oxy. In this case M1(x1';y1') and M2(x2';y2') have coordinates:

x1'=1; y1'=0; x2'=cos(α-β); y2'=sin(α-β)

this time distance between M1 and M2 is:

d2=(x2'-x1')2+(y2'-y1')2=(cos(α-β)-1)2+(sin(α-β)-0)2=cos2(α-β)-2cos(α-β)+1+sin2(α-β)=2-2cos(α-β)

let's equate the two results

2-2cosαcosβ-2sinαsinβ=2-2cos(α-β)

cos(α-β)=cosαcosβ+sinαsinβ

if we substitute β with -β we get

cos(α-(-β))=cosαcos(-β)+sinαsin(-β) and you hopefully know that cos is an even or w/e it's called function and sin is not, then

cos(α+β)=cosαcosβ-sinαsinβ

if you want the sine version just...

sin(α-β)=cos($\frac{\pi }{2}$-(α-β))=cos(($\frac{\pi }{2}$-α)+β)=cos($\frac{\pi }{2}$-α)cosβ-sin($\frac{\pi }{2}$-α)sinβ=sinαcosβ-cosαsinβ

and for the last one I will use the same old trick

sin(α-(-β))=sinαcos(-β)-cosαsin(-β)

sin(α+β)=sinαcosβ+cosαsinβ

And with these plus Pythagoras' theorem expressed with trig functions you could easily get tan and cotan and other important trig identities.

Note that I don't take into account angles from quadrants other than the first.

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14. May 28, 2012

### cool_jessica

Preliminaries

These can be seen from looking at the diagrams.
Sine and angle ratio identity

Proof: From the previous inequalities, we have, for small angles
, so
, so
, or
, so
, but
, so

Cosine and angle ratio identity

Proof:

The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero.
Cosine and square of angle ratio identity

Proof:
As in the preceding proof,

The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.
Proof of Compositions of trig and inverse trig functions
All these functions follow from the Pythagorean trigonometric identity. We can prove for instance the function

Proof:
We start from

Then we divide this equation by

Then use the substitution , also use the Pythagorean trigonometric identity:

Then we use the identity