Prove $\sin(-x) = -\sin(x)$
I'm looking for a really basic proof of $\sin(-x) = -\sin(x)$.
The proof should pretty much only employ basic trigonometry.
Thanks
$\endgroup$ 16 Answers
$\begingroup$$$ \sin(x) = \cos(\tfrac\pi2 - x) = \underbrace{\cos(\tfrac\pi2)}_{=0}\cos(-x) - \underbrace{\sin(\tfrac\pi2)}_{=1}\sin(-x) = -\sin(-x) $$
$\endgroup$ 6 $\begingroup$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...
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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}
$\endgroup$ 4 $\begingroup$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" $\cos(-\theta) \equiv \cos(\theta)$.
$\endgroup$ 5 $\begingroup$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)).
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