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I'm looking for a really basic proof of $\sin(-x) = -\sin(x)$.

The proof should pretty much only employ basic trigonometry.


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How do you define $\sin(x)$? The answer cannot be "opposite side over hypotenuse in a right triangle one with acute angle $x$," since then $\sin(90^\circ)$ or $\sin(-7)$ or ... are not defined. But if you do not specify what notion you are using, one cannot really verify anything. –  Andres Caicedo Jun 16 '14 at 0:30

5 Answers 5

up vote 12 down vote accepted

$$ \sin(x) = \cos(\tfrac\pi2 - x) = \underbrace{\cos(\tfrac\pi2)}_{=0}\cos(-x) - \underbrace{\sin(\tfrac\pi2)}_{=1}\sin(-x) = -\sin(-x) $$

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And how do we prove that $\sin(x)=\cos(\frac{\pi}2 -x)$? Or the formula for $\cos(a+b)$? –  Andres Caicedo Jun 16 '14 at 0:32
Brilliant! Just what I was looking for. –  Kermit the Hermit Jun 16 '14 at 0:34
@Midni Make sure you actually have a proof. Most elementary textbooks only prove the formulas for cosine or sine of a sum under additional assumptions (angles being acute, for instance), or may even appeal to the fact you want to prove, in which case you just went in circles and got nowhere. –  Andres Caicedo Jun 16 '14 at 0:38
@AndresCaicedo, I don't know what context you're assuming. The context I'm imagining is that we know trigonometry for angles in, say, $[0,\frac\pi2]$, and we'd like to understand how best to extend our definitions beyond this interval. This argument shows that, if the familiar identities are going to continue to hold outside that interval (which is surely what we want), then we're forced to define $\sin$ to be odd. If you think this isn't a good interpretation of the question, I'd like to hear what you have to say about it. –  Steven Taschuk Jun 16 '14 at 0:39
@AndresCaicedo I understand what you're saying. That question in the textbook was given after the proof of addition/subtraction identities of sin and cos was done. However the problem arises when wanting to use the proof of some of them because they accept that sin(-x) = -sin(x). Luckily in the case of Steven Taschuks' proof, using the subtraction formula for cos, there's no need to worry about any of that. Hope I have given you sufficient background regarding my post. –  Kermit the Hermit Jun 16 '14 at 0:51

A right-angled triangle with an incline of $x$ has height $\sin x$. A similar triangle with incline $-x$ has (signed) height $\sin(-x)$, and since this is just a reflection of our original triangle in the $x$-axis...

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Use the fact that $-x=0-x$ to get $$\sin(-x)=\sin(0-x)=\sin 0\cos x -\cos 0\sin x =0\cos x-1\sin x =-\sin x$$ as required.

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Here's a proof using complex exponentials:

$$e^{i\theta} = \cos\theta + i\sin\theta.$$

$$\overline{e^{i\theta}} = \overline{\cos\theta+i\sin\theta} = \cos\theta-i\sin\theta.$$

$$e^{-i\theta} = \cos(-\theta)+i\sin(-\theta).$$

It's not hard to see that $e^{i\theta}\overline{e^{i\theta}} = 1$ and also that $e^{i\theta}e^{-i\theta} = 1$. The first you can prove via Pythagorean theorem and the second you can prove by laws of exponentials. Due to uniqueness of inverses, $e^{-i\theta}$ must be the same as $\overline{e^{i\theta}}$ which in turn says that

$$ \cos\theta - i\sin\theta = \cos(-\theta)+i\sin(-\theta).$$

Equating real and imaginary parts gives

$$\cos\theta = \cos(-\theta)$$

and also

$$\sin(-\theta) = -\sin\theta.$$

To see that $e^{i\theta}\overline{e^{i\theta}} = 1$, note that this is nothing more than

\begin{eqnarray} (\cos\theta + i\sin\theta)\overline{(\cos\theta+i\sin\theta)} &=& (\cos\theta+i\sin\theta) (\cos\theta-i\sin\theta) \\ &=& \cos^2\theta+i\cos\theta\sin\theta-i\cos\theta\sin\theta-i^2\sin^2\theta \\ &=& \cos^2\theta+\sin^2\theta \\ &=& 1 \end{eqnarray}

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Does the overbar on your second line mean the complex conjugate? –  Cole Johnson Jun 23 '14 at 14:56
It most certainly does mean that. –  Cameron Williams Jun 23 '14 at 15:08
I'd expand a bit on the $e^{i\theta} \overline{e^{-i\theta}$. Mention that it expands to $\cos^2(\theta) + \sin^(\theta)$ which equals $1$ (by the Pythagorean Theorem). It took me a minute to realize that. –  Cole Johnson Jun 23 '14 at 16:47
I'd be happy to. Thanks for the suggestion. –  Cameron Williams Jun 23 '14 at 16:54

A visual approach to the problem is to look at a unit circle.

What is $\sin(\dfrac{\pi}{3})$? $\dfrac{\sqrt{3}}{2}$.

What is $\sin(\dfrac{\pi}{4})$? $\dfrac{\sqrt{2}}{2}$.

What is $\sin(\dfrac{\pi}{6})$? $\dfrac{1}{2}$.

What is $\sin(-\dfrac{\pi}{3})$? $-\dfrac{\sqrt{3}}{2}$.

What is $\sin(-\dfrac{\pi}{4})$? $-\dfrac{\sqrt{2}}{2}$.

What is $\sin(-\dfrac{\pi}{6})$? $-\dfrac{1}{2}$.

Notice a pattern? This approach can also be used to "prove" $\cos(-\theta) \equiv \cos(\theta)$.

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This indeed supports, but doesn't prove, the hypothesis. Still, I agree that it is quite important to believe that a that a formula is correct, and to see supporting examples, before accepting its truth. –  MPW Jun 23 '14 at 2:03
The danger is: False Claim: $\sin(-x)=\sin x$. Supporting Examples of False Claim: $\sin (-\pi k)=\sin\pi k$ for integral $k$. –  MPW Jun 23 '14 at 2:07
@MPW that may be true. I never meant for one to use this approach exclusively. I just meant it to be a nice way to visualize why. –  Cole Johnson Jun 23 '14 at 2:33
Understood. I was actually thinking of something along these lines myself. It does seem to occur naturally as a reasonable illustration, doesn't it? –  MPW Jun 23 '14 at 7:49
@MPW yep. However, in response to your original comment, I put the word "prove" in quotes. –  Cole Johnson Jun 23 '14 at 14:55

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