I have been trying to solve this problem:
"Use sum and difference identities and the sine and cosine functions to deduce that:"
$$\tan(\pi – x) = -\tan(x)$$
I can see that:
$$\tan(\pi – x) = \frac{\sin(\pi – x)}{\cos(\pi – x)} = \frac{\sin(\pi)\cos(x) – \sin(x)\cos(\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)}$$
But I have no idea where to go from here. I can't solve this algebra after a long time and really need to ask where to from here.
Best Answer
Since $\sin(\pi)=0, \cos(\pi) = -1$ you have $$\tan(\pi - x) = \frac{\sin(\pi - x)}{\cos(\pi - x)} = \frac{\sin(\pi)\cos(x) - \sin(x)\cos(\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)} = \frac{+\sin(x)}{-\cos(x)} = -\frac{\sin(x)}{\cos(x)} = -\tan(x)$$