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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.

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    $\begingroup$ Hint: $\sin(\pi)=0$ and $\cos(\pi)=-1$. $\endgroup$
    – Nizar
    Commented Nov 3, 2015 at 13:32
  • $\begingroup$ That's done it! $\endgroup$
    – Hanshan
    Commented Nov 3, 2015 at 13:37
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    $\begingroup$ I beg you pardon, but $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$ for all $x$ are basic facts about sine and cosine. $\endgroup$
    – Bernard
    Commented Nov 3, 2015 at 13:38
  • $\begingroup$ True, @Bernard. However, the OP was instructed to do it using the sine and cosine sum and difference identities. I guess they could do it by showing your identities as an intermediate step, though. $\endgroup$
    – PM 2Ring
    Commented Nov 3, 2015 at 13:41

3 Answers 3

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Since $\sin(\pi)=0, \cos(\pi) = -1$ you have $$\tan(\pi - x) = \frac{\sin(\pi - x)}{\cos(\pi - x)} = \frac{\sin(\pi)\cos(x) - \sin(x)\cos(\pi)}{\cos(\pi)\cos(x) + \sin(\pi)\sin(x)} = \frac{+\sin(x)}{-\cos(x)} = -\frac{\sin(x)}{\cos(x)} = -\tan(x)$$

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  • $\begingroup$ Thank you, I can now see that very clearly. $\endgroup$
    – Hanshan
    Commented Nov 3, 2015 at 13:38
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Hint

$\tan(A-B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$. Use $A = \pi$ and $B = x$

So $\frac{\tan \pi-\tan x}{1 + \tan π\tan x} =-\tan x$ . (Since $\tan \pi = 0$).

So $\tan (\pi - x) = -\tan x$ (proved)

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  • $\begingroup$ You have to learn to format your answers. Otherwise they will get downvoted. You have been there on this site for a while now $\endgroup$
    – Shailesh
    Commented Nov 3, 2015 at 13:46
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    $\begingroup$ ya sure will try to improve. $\endgroup$
    – Archis Welankar
    Commented Nov 3, 2015 at 13:58
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    $\begingroup$ Also note that you had the formula wrong. The denominator has a + sign $\endgroup$
    – Shailesh
    Commented Nov 3, 2015 at 14:18
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Hint:
You can use this property, $$ \cos(\pi-x)=-\cos x $$ Or by using this identity, $$ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} $$ Substitute $A$ for $\pi$ and B for $x$

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