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Can someone explain to me why $\sin\left(\frac xy\right)$ is not equal to $\frac{\sin x}{\sin y}$, and as an extension, why this holds true for all trig functions? Also, why is $\sin(x+y)$ is not equal to $\sin x+\sin y$, and why does this holds true for all trig functions?

I get that this may be because they are functions, but what about the nature of trig functions causes the two to examples above to be not equal?

By the way, can you please keep the explanation very simple please? I am a high school student and may struggle to understand more complex explanations involving proof notation etc.

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  • $\begingroup$ why is sin(x+y) is not equal to sin(x)+sin(y) See Overview of basic facts about Cauchy functional equation. $\endgroup$ – dxiv Jul 19 '18 at 6:55
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    $\begingroup$ There is no reason why two expressions should give the same value just because they "kinda sorta look similar". Do you have any reason to think that $\sin(x+y)$ should be the same as $\sin x+\sin y$? Do you have any difficulty understanding that $(x+y)^2$ is not the same as $x^2+y^2$? $\endgroup$ – David Jul 19 '18 at 7:00
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    $\begingroup$ Do you know that $-1\leq \sin(x)\leq 1$ for all $x\in \mathbb{R}$? With this fact, it is really easy to see why your equalities cannot hold. $\endgroup$ – Lorenzo Quarisa Jul 19 '18 at 7:07
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    $\begingroup$ Why on earth should $\sin(x+y)$ be anything like $\sin x+\sin y$. I mean, the graph of $\sin$ is not a straight line is it? $\endgroup$ – Lord Shark the Unknown Jul 19 '18 at 7:57
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    $\begingroup$ it's best to try and understand them with geometry perhaps - math10.com/en/geometry/trigonometry-and-geometry-conversions/… $\endgroup$ – Cato Jul 19 '18 at 9:28
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Just assume $x=y\neq 0$ then

$$\sin(x/x)=\sin 1\neq \sin x/\sin x=1$$

and

$$\sin(x+x) =\sin (2x)=2\sin x\cos x\neq \sin x +\sin x=2\sin x$$

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    $\begingroup$ Does this tell "why" ? $\endgroup$ – Yves Daoust Jul 19 '18 at 7:54
  • $\begingroup$ @YvesDaoust It depends how we consider the question "why?". To keep the explanation very simple, I've given some simple counterexamples to see "why it is not true" and we can find many others of them. $\endgroup$ – user Jul 19 '18 at 7:59
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    $\begingroup$ Why does it rain today ? Because the ground is wet. $\endgroup$ – Yves Daoust Jul 19 '18 at 8:05
  • $\begingroup$ @gimusi I largely get your answer, but how do we know that what you've written above is true even when y is not the same as x? $\endgroup$ – Ethan Chan Jul 19 '18 at 8:15
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    $\begingroup$ @EthanChan What we show by a counterexample is that the relations are not true in general and thus that they are false. Of course we can find some special cases for which they are true. $\endgroup$ – user Jul 19 '18 at 8:17
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A second of thinking tells you that no function achieves that !

$$\sqrt{x+y}\ne\sqrt x+\sqrt y,\frac1{x+y}\ne\frac1x+\frac1y,\log(x+y)\ne\log x+\log x,\cdots$$

$$\frac xy+1\ne\frac{x+1}{y+1},\tan\frac xy\ne\frac{\tan x}{\tan y}\cdots$$


There are just two exceptions:

$$a(x+y)=ax+ay$$

and

$$\left(\frac xy\right)^a=\frac{x^a}{y^a}.$$

So you'd better ask why the linear function is additive and why the power function is multiplicative.


If you want to find all additive functions, i.e. such that

$$f(x+y)=f(x)+f(y),$$ you immediately see that

$$f(2x)=2f(x)$$ and by induction

$$f(nx)=nf(x).$$

This generalizes to rationals,

$$qf\left(\frac pqx\right)=qpf\left(\frac xq\right)=pf(x)\implies f\left(\frac pqx\right)=\frac pqf(x),$$ and to reals

$$f(rx)=rf(x),$$ but the proof is more technical.

Now, setting $r\to x,x\to1$,

$$f(x)=f(1)\,x=ax.$$


For the multiplicative functions

$$g(xy)=g(x)g(y)$$

consider the function

$$f(x):=\log g(e^x)$$ and observe that it is additive, so that

$$f(x)=\log g(e^x)=ax,$$

$$g(e^x)=e^{ax},$$

$$g(x)=x^a.$$

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  • $\begingroup$ Moreover, the only function satisfying both conditions at once is the identity. $\endgroup$ – TonioElGringo Jul 19 '18 at 8:16
  • $\begingroup$ The extension to reals only works if the function is continuous. $\endgroup$ – Arnaud D. Jul 19 '18 at 9:06
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If you are looking at the universal statement:

$\sin(x/y)$ is not equal to $\sin(x)/\sin(y)$, for all $x$ and $y$

This statement is false.

Let $x=\pi^2/(\pi + 2)$, $y=\pi/2$: $$x/y=\pi-x$$ $$\sin(x/y)=\sin(x)$$ $$\sin(y)=1$$ $$\therefore \sin(x/y)=\sin(x)/\sin(y)$$

The universal statement:

$\sin(x+y)$ is not equal to $\sin(x)+\sin(y)$, for all $x$ and $y$

is also false, since $\sin(2\pi)+\sin(4\pi)=\sin(2\pi+4\pi)$.

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Note that when $y=\pi$, $$\sin(x+y)=\sin(x+\pi)=-\sin x$$ but $$\sin x+\sin y=\sin x+\sin \pi=\sin x.$$


Also, when $x=2y$, $$\sin\left(\frac xy\right)=\sin2$$ whch is a constant for all $x,y\neq0$ but $$\frac{\sin x}{\sin y}=\frac{\sin 2y}{\sin y}=\frac{2\sin y\cos y}{\sin y}=2\cos y$$ which varies as $y$ varies.

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Observe that

$$\sin(x+y)=2\sin\frac{x+y}2\cos\frac{x+y}2$$ and $$\sin(x)+\sin(y)=2\sin\frac{x+y}2\cos\frac{x-y}2.$$

The two expressions are equal when $$x+y=2k\pi$$ or when

$$\cos\frac{x+y}2=\cos\frac{x-y}2,$$

or

$$x+y=\pm(x-y)+4k\pi$$

which is true with

$$x=2k\pi\text{ or }y=2k\pi.$$

Hence $$\sin(x+y)=\sin x+\sin y$$ on a grid of step $2\pi$ plus all NW-SE diagonals.


The case of the division is much less tractable.

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I think the other answers don't really address the heart of the question, and are too focused on proving why @Ethan_Chan's equations are false specifically for trigonometric functions. Furthermore, I think @Ethan_Chan's misunderstanding is on a more fundamental level.

$\sin(x)$ is a function, like any other. But why shouldn't every function satisfy $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ or $f(x+y)=f(x)+f(y)$? The reason is that a function can map different inputs ($x$ and $y$) to different outputs ($f(x)$ and $f(y)$). But the relationships between $x$ and $y$ aren't necessarily the same as the relationships between $f(x)$ and $f(y)$.

Let's think about what's happening to $x$ and $y$ in each of your examples. You're performing an operation on them: division ($\frac{x}{y}$) or addition ($x+y$). Let's denote an operation on $x$ and $y$ as $x*y$ (be it addition, subtraction, etc.). This brings us to a general rule for why your examples don't work:

$$\text{in general, }f(x*y)\neq f(x)*f(y)$$

@Yves_Daoust's answer shows us the exceptions to this rule. But in general, it is true.

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