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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.

Thanks

Paul
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Kermit the Hermit
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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. – Andrés E. Caicedo Jun 16 '14 at 00:30

5 Answers5

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

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    And how do we prove that $\sin(x)=\cos(\frac{\pi}2 -x)$? Or the formula for $\cos(a+b)$? – Andrés E. Caicedo Jun 16 '14 at 00:32
  • Brilliant! Just what I was looking for. – Kermit the Hermit Jun 16 '14 at 00:34
  • @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. – Andrés E. Caicedo Jun 16 '14 at 00:38
  • @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. –  Jun 16 '14 at 00:39
  • "I don't know what context you're assuming." I am not assuming any context, actually. I'm asking the poster to indicate what context they are assuming. – Andrés E. Caicedo Jun 16 '14 at 00:43
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    @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 00:51
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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.

MPW
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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...

preferred_anon
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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}

Cameron Williams
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Easy, think about the construction of the Unit Circle. We can simply make an observation of this. Since -x is the same angle as x reflected across the x-axis, sin(-x) =-sin(x) as sin(-x) reverses it's positive and negative halves sequentially when you think of the coordinates of points on the circumference of the circle in the form p = (cos(x),sin(x)).

run_tron
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