Prove $\sin(-x) = -\sin(x)$

Question

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

The proof should pretty much only employ basic trigonometry.

Thanks

Answer 1

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

Answer 2

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...

Answer 3

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

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

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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}

Answer 4

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.

Answer 5

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)).