Sine Angle Product Formula 2

In this post, we continue the discussion from a previous post about the existence of a suitable sine angle product identity.

Let’s try a different approach. We know that $\sin a$ and $\cos a$ are transcendental for any nonzero algebraic number $a$ . We also know that we can express $\sin{na}$ as an $n$ -degree polynomial in $\sin a$ and $\cos a$ for $n \in \mathbb{Z}_{+}$ (look at Wikipedia’s list of trig identities.) Can we show that, with $n$ fixed and $a$ varying, any polynomial expression for $\sin{na}$ must be at least degree $n$ ? Can we use this then for our $\sin{ab}$ question and use infinite degree contradiction?

—

Say $n$ is fixed, and $P$ is a polynomial such that

$\displaystyle \sin{na} = P\left( \sin a \right)\ for\ all\ a.$

Now, say there existed some other polynomial $Q$ such that

$\displaystyle \sin{na} = Q\left( \sin a \right)\ for\ all\ a.$

Then,

$\displaystyle P\left( \sin a \right) - Q\left( \sin a \right) = 0\ for\ all\ a.$

But we can choose infinitely many values of $a$ with distinct values of $\sin a$ (the sine function goes continuously from -1 to 1), thus the Factor Theorem guarantees this is impossible.

Note however that this does not directly show that the polynomial identity for $\sin{na}$ given by Wikipedia is unique, for that identity involves $\cos a$ as well as $\sin a$ .

Can we adapt this to show uniqueness of a polynomial identity for $\sin{ab}$ ? Let’s say there existed two distinct polynomials $P,Q$ such that, for all $a,b$ ,

$\displaystyle \sin{ab} = P\left( \sin a,\cos a,\sin b,\cos b \right) = Q\left( \sin a,\cos a,\sin b,\cos b \right).$

Actually, this is not unique: we can in fact write a counterexample, $Q(r,s,t,u) = P(r,s,t,u) + \left( r^{2} + s^{2} - 1 \right)$ . This further shows that an entire infinite family of polynomial expressions can replace $P$ , $Q(r,s,t,u) = P(r,s,t,u) + c\left( r^{2} + s^{2} - 1 \right)$ for any $c$ . Thus, $P$ cannot be unique.

—

Maybe we can yield some progress by removing cosines and dealing only with $\sin a,\sin b$ . Say there existed a polynomial $P$ such that, for all $a,b$ ,

$\displaystyle \sin{ab} = P\left( \sin a,\cos a,\sin b,\cos b \right).$

We try to show that there exists a polynomial $Q$ such that, for all $a,b$ ,

$\displaystyle \sin{ab} = Q\left( \sin a,\sin b \right)$

(with cosines removed.)

Unfortunately, we can shed some doubt on our ability to do this. We disprove the following: given a two-variable polynomial $P$ , there exists a single-variable polynomial $Q$ such that $Q\left( \sin a \right) = P\left( \sin a,\cos a \right)$ for all $a$ . Indeed, consider $P(x,y) = x + y$ , and assume for the sake of contradiction $Q$ existed. So we want $Q\left( \sin a \right) = \sin a + \cos a$ . We can write this as $(Q - I)\left( \sin a \right) = \cos a$ , where $I$ is the identity $I(x) = x$ . Thus, $\left( (Q - I)\left( \sin a \right) \right)^{2} = \cos^{2}a = 1 - \sin^{2}a$ , so there is a (nonzero) polynomial $R$ such that $R\left( \sin a \right) = 0$ for all $a$ . But we can choose infinitely many $a$ with distinct values of $\sin a$ , yielding a contradiction.

We can repeat this proof to show the non-existence of $Q$ for any $P$ of the form $P(x,y) = A(x) + y$ , where $A$ is any single-variable polynomial. In fact, we have non-existence of $Q$ for any $P$ of the form $P(x,y) = A(x) + yB(x)$ , where $A,B$ are any single-variable polynomials.

—

Can we show the following? For $n \in \mathbb{Z}_{+}$ , if there exists a two-variable polynomial $P$ such that $\sin{na} = P\left( \sin a,\cos a \right)$ for all $a$ , then $P$ must be at least degree $n$ . Would that yield our desired result?

Let’s assume this is the case. Assume we had a polynomial $P$ such that for all $a,b$ ,

$\displaystyle \sin{ab} = P\left( \sin a,\cos a,\sin b,\cos b \right).$

Let $n$ be the degree of $P$ . But now let $b = n + 1$ . Let the two-variable polynomial $P^{'}$ be defined by $P^{'}(x,y) = P\left( x,y,\sin(n + 1),\cos(n + 1) \right)$ . Then, $P^{'}$ is at most degree $n$ by definition. But we have

$\displaystyle \sin\left( (n + 1)a \right) = P^{'}\left( \sin a,\cos a \right)\ for\ all\ a,$

so $P^{'}$ must have degree at least $n + 1$ . This is a contradiction.

We can yield a similar result if we show that the polynomial expressions for $\sin{na}$ must have degree tending to infinity as $n$ tends to infinity. Thus, if we show that, we’re done.

We can attempt this by induction!

—

Toy example: $\sin{2a} = 2\sin a\cos a$ . Why does this have to be degree 2? Well, we have

$\displaystyle \sin(2a) = \sin(a + a) = \sin a\cos a + \cos a\sin a.$

Each of these terms increases the degree by 1. Thus, if the best we can do for sine and cosine is degree $n$ , then the best we can do for sine and cosine for $n + 1$ is degree $n + 1$ .

Is it the best though? We’re going one way, not the other.

OK, let’s say then that for the sake of contradiction there was a linear expression for $\sin(2a)$ in terms of $\sin a,\cos a$ . In other words, say that there were constants $c,d,e$ such that

$\displaystyle \sin(2a) = c\sin a + d\cos a + e$

for all $a$ . To mimic our proof structure, we have that

$\displaystyle \sin a\cos a + \cos a\sin a = c\sin a + d\cos a + e,$

and we’re trying to show this isn’t possible.

So how can we do this? We’ve already used all the trigonometry that we think is applicable, with using the sine angle addition formula. Equivalently, we want to show the impossibility of

$\displaystyle \sin a\cos a + \cos a\sin a - c\sin a - d\cos a - e = 0.$

Why is this so impossible? Well, we know in this case that the sinusoidal coefficient of $a$ cannot match, so these functions will be out of phase and cannot be equal for all $a$ . But that involves specific info around $n = 2$ , so how can we do a generalized argument?

We know that it’s impossible for $\sin a$ to be a constant. That’s basically the induction step assumption for this toy case. How do we go from that to yielding the impossibility of this?

—

Let’s try $n = 3$ , still concrete and a toy example, but maybe could give us more insight into how the induction step assumption can be used. We want to prove the impossibility of: there is a quadratic $P$ such that

$\displaystyle \sin(3a) = P\left( \sin a,\cos a \right).$

We know that there cannot be a linear expression for $\sin{2a}$ or $\cos{2a}$ .

So we have

$\displaystyle \sin(2a + a) = \sin{2a}\cos a + \cos{2a}\sin a$

$\displaystyle = P\left( \sin a,\cos a \right).$

Thus,

$\displaystyle \sin{2a}\cos a = P\left( \sin a,\cos a \right) - \cos{2a}\sin a.$

How do we proceed?

—

Maybe it’s about the phase.

Consider the function

$\displaystyle f(x) = \sum_{i,j = 0,i + j \leq n}^{}{a_{ij}\sin^{i}x\cos^{j}x}.$

Can we write $f$ as a sinusoidal function, or can we show that it cannot have phase coefficient $n + 1$ ? If that’s the case, we’re done.

I think I know how to do this. It’s just about writing out the details now.

Toy example:

$\displaystyle \sin^{3}x\cos^{4}x = \left( \sin x\cos x \right)^{3}\cos x = \left( \sin{2x} \right)^{3}\cos x$

$\displaystyle = \left( \sin{2x} \right)^{2}\left( \sin{2x}\cos x \right)$

$\displaystyle = \left( \sin{2x} \right)^{2}\left( \sin(3x) + \sin x \right)$

$\displaystyle = \left( \sin{2x} \right)^{2}\sin x + \left( \sin{2x} \right)^{2}\sin(3x)$

$\displaystyle \ldots$

$\displaystyle \sin(a + b) + \sin(a - b) = 2\sin a\cos b$

$\displaystyle \ldots$

It remains to continue this.

—

Can we show that if there is a polynomial expression between $\sin a$ and $\cos a$ , it must have degree at least 2? (We know $P(x,y) = x^{2} + y^{2} - 1$ works.)

Yes, we can! Let’s say there was a linear expression

$\displaystyle c\sin a + d\cos a + e = 0.$

It’s easy to show from this that $c = d = e = 0$ , just from the sinusoidal forms.

So this shows the following: if we have two different polynomial expressions $P,Q$ for $\sin{na}$ in terms of $\sin a$ and $\cos a$ , then their difference must be a polynomial of degree at least 2.

Can we show something even stronger \ldots{} that the difference must have $P(x,y) = x^{2} + y^{2} - 1$ as a factor? If that’s the case, then does that yield our result?

We know that an $n$ -degree expression exists for $\sin{na}$ . Let’s say for the sake of contradiction that there was also an expression of degree less than $n$ . The difference between these two expressions has degree $n$ and must have $x^{2} + y^{2} - 1$ as a factor. Thus, it can be written

$\displaystyle P(x,y) - Q(x,y) = A(x,y)\left( x^{2} + y^{2} - 1 \right).$

It remains to continue this.

—

Let’s write out the induction approach. We attempt to show a generalization: any polynomial expression involving $\sin{na},\cos{na},\sin a,\cos a$ must have degree at least $n$ . This is sufficient for our desired application, and it may allow us more flexibility in the induction step.

Base case: $n = 1$ . Say there is a polynomial $P$ such that $P\left( \sin a,\cos a,\sin a,\cos a \right) = 0$ for all $a$ . If $P$ doesn’t have degree at least 1, then it is nonzero constant, but this is impossible, contradiction.

Toy example: $n = 2$ . Say there is a linear polynomial $P$ such that $P\left( \sin{2a},\cos{2a},\sin a,\cos a \right) = 0$ . We have

$\displaystyle P\left( 2\sin a\cos a,\cos^{2}a - \sin^{2}a,\sin a,\cos a \right) = 0.$

But how do we proceed now?

Induction step: now, assume the result for $n$ ; we prove it for $n + 1$ . Assume for contradiction there was a polynomial $P$ of degree at most $n$ such that, for all $a$ ,

$\displaystyle P\left( \sin{(n + 1)a},\cos{(n + 1)a},\sin a,\cos a \right) = 0.$

We have

$\displaystyle \sin{(n + 1)a} = \sin(na + a) = \sin{na}\cos a + \cos{na}\sin a,$

$\displaystyle \cos{(n + 1)a} = \cos(na + a) = \cos{na}\cos a - \sin{na}\sin a,$

thus

$\displaystyle P\left( \sin{na}\cos a + \cos{na}\sin a,\cos{na}\cos a - \sin{na}\sin a,\sin a,\cos a \right) = 0.$

But now this is a degree at most $n$ expression for $\sin{(n + 1)a}$ in terms of $\sin a,\cos a$ :

$\displaystyle Q\left( \sin a,\cos a \right) = \sin\left( (n + 1)a \right) = \sin(na + a) = \sin{na}\cos a + \cos{na}\sin a.$

Thus, for all $a$ ,

$\displaystyle \sin{na}\cos a = Q\left( \sin a,\cos a \right) - \cos{na}\sin a.$

It remains to continue this.

Written December 12, 2022.

—

This discussion is continued in a follow-up post.

Share and like (need WordPress account to like):

Leave a comment Cancel reply