Proving tan(x)=-tan(-x)

By Hal 2001 (P3046) on Thursday, November 30, 2000 - 07:27 pm :

Hello all,

How do you go about showing that tan(x) = -tan(-x)
Can you show it using a unit circle, by drawing a tangent to the circle at right angles to the radius??

By Dan Goodman (Dfmg2) on Thursday, November 30, 2000 - 08:58 pm :

Well, tan(x)=sin(x)/cos(x), so I would have thought the easiest way to do it would be to say tan(-x)=sin(-x)/cos(-x)=(-sin(x))/cos(x)=-tan(x). Are you happy with sin(-x)=-sin(x) and cos(-x)=cos(x)?

By Kerwin Hui (Kwkh2) on Friday, December 1, 2000 - 12:03 pm :

Another method is as follows (provided, of course,

\cos θ \neq 0

Draw a line through origin making an angle $θ$ with the $x$ -axis. Extend this line to meet the line $x = 1$ . The $y$ -coordinate is now $\tan θ$ .

Similarly, draw another line through origin making an angle $- θ$ with the $x$ -axis. Extend this line to meet the line $x$ =1.

It is then quite obvious from geometry that $\tan θ = - \tan (- θ)$ .

Kerwin

By Hal 2001 (P3046) on Friday, December 1, 2000 - 01:49 pm :

Dan, yes, I agree that cos(x)=cos(-x) {reflection in the y axis, right?} sin(-x)=-sin(x) {is this a reflection in the y axis then reflected in x axis, is 180deg turn, right?} Your method is neat and simple. Thanks.

Kerwin, your method is sort of using the 4 quadrants, is that right? This is what I tried to describe in the original post. I tried to describe it using a circle. But the words did not come out properly. I like both methods.

Thanks Nrich Team!
Hal2001