I've read somewhere that the sine and cosine functions can be fully described by this theorem:
- $\sin(0) = 0, \cos(0) = 1$
- $\sin(a-b) = \sin(a)\cos(b) – \sin(b)\cos(a)$
- $\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$
- There is na $r>0$ such that:
$$0<\sin(x)<x<\tan(x), x \in(0, r), \tan(x) = \frac{\sin(x)}{\cos(x)} $$
With this theorem, we can prove things like:
-
$\sin^2(x) + \cos^2(x) = 1$
by doing$\cos(a-a) = \cos(a)\cos(a) + \sin(a)\sin(a)$ -
$\sin(-x) = -\sin(x)$
by doing$\sin(0-x) = \sin(0)\cos(x) – \sin(x)\cos(0)$ - $\cos(a+b) = \cos(a)\cos(b) – \sin(a)\sin(b)$
by doing$\cos(a-(-b)) = \cos(a)\cos(-b) + \sin(a)\sin(-b)$
- $\sin(a+b) = \sin(a)\cos(b)+\sin(b)\cos(a)$
by doing$\sin(a-(-b)) = \sin(a)\cos(-b) – \sin(-b)\cos(a)$
And other trigonometric identities that also follows from what I've already done.
The problem is that there are many definitions for the sine and cosine function. Let's begin by the classical definition:
Classical Definition
The sine function is defined as the ratio between the opposite side of the angle, and the hipotenuse of this right triangle.
The cosine function is defined as the ratio between the adjacent side of the angle, and the hipotenuse of this right triangle.
The tangent function is defined as the ratio between the sine function and the cosine function (with $\cos (x) \neq 0)$
The other trigonometric identities can be proven geometrically for an angle less than or equal $\frac{\pi}{2}rad$ because it's a right triangle. So, we can't prove $\sin(a-b)$ geometrically and then prove $\sin(a+b)$ analytically like I did, because we assumed a negative $b$, something that's not defined geometrically in the right triangle.
The identity $\sin^2(x) + \cos^2(x) = 1$ can be proven with a simple pythagorean theorem in a triangle with hypotenuse 1.
The formulas for the sum and difference of sines and cosines can be proven geometrically like in this images I found in this answer:
Unit circle definition
Imagine a circle centered at the origin of the cartesian plane, then:
The sine function for an $x \in R, x>0$ can be defined as the $y$ position of point of the circle where the angle stops if we travel anti-clockwise inside the circle's line.
The cosine for an $x \in R, x>0$ function can be defined as the $x$ position of point of the circle where the angle stops if we travel anti-clockwise inside the circle's line.
We can make the same definition for negative angles, so for an $x \in R, x<0$ the same holds, but we're know travelling clockwise.
The tangent function is defined as the ratio between the sine function and the cosine function (with $\cos (x) \neq 0)$
Then, we can define these functions for all real numbers, since when we travel $2\pi$ we get back to the initial point. So we defined sine and cosine as periodic functions.
Open question: How do I prove, with the unit circle periodic definition and without being circular, the $\sin(a-b)$ and $\cos(a-b)$ formulas? (same for $\sin(a+b)$ and $\cos(a+b)$.
Taylor Series definition
$$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$$
$$\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}$$
$$\tan x = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1}$$
Here, in the same question, there's an analytic proof of the trigonometric identites, for these sums.
What's the best definition for calculus?
Well, in calculus we use trigonometric functions a LOT: in integral substitutions, in series, taylor series (like the ones I showed now), derivatives, convergence tests (like the Euler onde in the basel problem) and other things…
All the definitions I see, are kinda circular or not rigorous enough to make me feel good taking some derivatives or integral substitution, because I always care about the domain of these things. So I want to define it very nicely and be able to use all the trigonometric identities.
I've seen many geometrical proofs of $\sin(a-b)$, $\sin(a+b)$, $\cos(a-b)$, $\cos(a+b)$ using a right triangle and then suddenly the person starts using this formula for all real numbers. I need a complete definition of the trigonometric functions that works periodically and for all reals. The taylor series definition seems good but they're generated using trigonometric identities that are not yet proven (assuming this definition).
ps: I know that I used some primitive words in some definitions, like 'travel' so I let them in emphasis, but I hope you guys understand. And sorry by the long post, but I needed to do it, because I've never seen a complete definition in any book. Thanks.
Best Answer
The standard definition of the trig functions is in terms of a unit circle. The classical definition in terms of right triangles is more limited in terms of scope. For this reason in Calculus the more robust definition in terms of the unit circle is used. From this 'standard' definition every identity about the trigonometric functions can be proven.
In your post above you gave the definition in terms of the unit circle, but you left what I think is an important point. When defined in terms of the unit circle the trig functions are 'real valued'. What we mean here is that they are defined in terms of a distance and not some constructed idea called an angle. So let me begin with the definition in terms of the unit circle, then we will move to the sister proof to the one given above for the sine of a difference. From this we can establish the other three identities.
Consider the unit circle $$ \begin{equation}x^2+y^2 = 1 \end{equation} $$ Starting at the x-intercept (1,0) we travel a distance t around the circle in a counter-clockwise direction (a clockwise movement would be assigned a negative sign). Once we travel a distance t around the circle we end at a point P(x,y) on the circle. We then define $$ \begin{eqnarray} \cos t = x \\ \sin t = y \end{eqnarray} $$
It then turns out that we can return to angles by noting that on the unit circle the radian measure of the central angle subtended by the arc of length t is equal to t. But we don't need to ever actually return to angles. Our definition stands on its own. Now an immediate consequence of our definition is the familiar Pythagorean identity for sine and cosine.
$$\begin{equation} \cos^2 t + \sin^2 t = 1 \end{equation} $$
(This is obtained by substituting sine and cosine into the unit circle equation for x and y.)
Before we move onto the identity you asked about first let us notice that the sine and cosine functions are periodic. IE. $$ \cos(a + 2k \pi) = \cos a $$. I believe this is pretty clear because any trip around the circle will take you back to where you started. Make that trip k times and you are still back where you started. For this reason as mentioned above we can assume both angles are in$$ [0, 2 \pi)$$. Also note that $$\cos (-t) = \cos t $$ for all real numbers t.
The last thing to address is that rotation is a rigid transformation. In other words when we rotate the plane about a point, the distance between two points before rotation is the same after rotation.
Now we return to the problem. Show that
$$\cos(a-b) = \cos a \cos b + \sin a \sin b $$
Without loss of generality assume $$0<b<a<2 \pi $$ (the case of b=0 is trivial, as is the case of a = 2 $\pi$, and the case a=b, you can verify these yourself)
We now consider the points (see diagrams) P($\cos a, \sin b)$ and Q($\cos b, \sin b)$ on the unit circle. We can use the distance formula to find the length of segment PQ. We then rotate the entire plane clockwise through a distance b along the circumference of the circle (or through an angle with radian measure b). This will place point Q at Q`(1,0) and P will now be a distance b-a along the circle from (1,0) so it will be at P`($\cos(a-b), \sin(a-b))$. We now proceed to use the distance formula. Since distance is preserved through rotation segment PQ and segment P`Q` will have the same length.
$$\begin{eqnarray} \sqrt{(\cos a - \cos b)^2 + (\sin a - \sin b)^2} = \sqrt{(\cos(a-b)-1)^2 + (\sin(a-b) - 0)^2} \\ \cos^2 a - 2 \cos a \cos b + \cos^2 b + \sin^2 a - 2 \sin a \sin b + \sin^2 b = \cos^2(a-b) - 2 \cos(a-b) + 1 + \sin^2(a-b) \\ \cos^2 a + \sin^2 a + \cos^2 b + \sin^2 b - 2 \cos a \cos b - 2 \sin a \sin b = \cos^2(a-b) + \sin^2(a-b) + 1 - 2 \cos(a-b) \\ - 2 \cos a \cos b - 2 \sin a \sin b = - 2 \cos(a-b) \\ \cos a \cos b \sin a \sin b = \cos(a-b) \end{eqnarray} $$
Now I assumed $b<a$, is that ok? The answer is yes but you might want to answer why.
Armed with the cosine of the difference formula we can now prove the following.
$$ \begin{eqnarray} \cos(a+b) = \cos(a-(-b)) = \cos a \cos(-b) - \sin a \sin(-b) \\ = \cos a \cos b - \sin a \sin b \end{eqnarray} $$
The last step in the equality comes from our identity for $\cos(-t)$ above and the fact that $\sin(-t) = -\sin(t) $ as was mentioned in the previous post.
Now we can show that $\cos(\frac{\pi}{2} - x) = \sin x$ and in a similar manner to the problem above that $\sin(\frac{\pi}{2} - x) = cos x$. These are your familiar co-function identities. Finally armed with these equations we can now demonstrate the sine of a difference which can then prove the sine of a sum formula.
$$ \begin{eqnarray} \sin(a-b) = \cos(\frac{\pi}{2} - (a-b)) \\ = \cos((\frac{\pi}{2} -a) + b) \\ = \cos(\frac{\pi}{2} -a) \cos b - \sin(\frac{\pi}{2} -a) \sin b \\ = \sin a \cos b - \cos a \sin b \end{eqnarray} $$
And so there we have it. I apologize if the formating is off a bit, still getting used to the system. But now from this identities you can prove most any other basic identity involving the trig functions. All of your double and half angle formulas and your sum-product and product-sum formulas.