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Here's a piture of what I'm trying to say

I can't get my head around this, any help would be very much appreciated. Thanks

EDIT: t is an angle, where 0 < t < 90, angle t is in degrees EDIT: Added a picture I lifted from google

seeker
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    Please clarify: What is $t$? – inZugzwang Mar 25 '13 at 12:48
  • an angle, 0<t<90 – seeker Mar 25 '13 at 12:51
  • angle in radians or degrees? – jimjim Mar 25 '13 at 12:53
  • It's in degrees – seeker Mar 25 '13 at 13:00
  • @Assad: Are you saying that the graph of $f(t) = \tan(t)$ touches the unit circle at $(1,0)$? I don't understand the question. – Thomas Mar 25 '13 at 13:13
  • What do you mean by "touch the unit circle"? $\tan(t)$ when $0^\circ < t < 90^\circ$ is any positive number, it's not a point in the plane. – Yoni Rozenshein Mar 25 '13 at 13:14
  • I'm sorry, I've seen to have made a mess of describing myself, I'm pretty sure you've all seen the trig functions on a unit circle, where cos(t)=x and sin(t)=y, well I've also seen tan(t) on this unit circle and was wondering why it is a tangent to the unit circle at (1,0), why that specific point? – seeker Mar 25 '13 at 13:17
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    For convenience. You could well define another function, puting this line anywhere, but it wouldn't give all these pretty formulas. By the way, if your line touches $(0,1)$, then you get cotangent. – Jean-Claude Arbaut Mar 25 '13 at 13:52

3 Answers3

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I think what you're trying to express is that the length of the vertical segment joining $P=(1,0)$ to the point labelled $R$ in the diagram is the value of $\tan h$ where $h$ is the angle in the diagram. When $R$ is above the $x$ axis (as in the diagram), this is true by using the definition of $\tan h$ as "opposite over adjacent", because in this picture the "adjacent" has length $1$, being the segment from $O=(0,0)$ to $P=(1,0).$

The same thing works if $R$ happens to be below the $x$ axis, provided we interpret the segment joining $P$ to $R$ as having negative length (i.e. "signed length" is negative) because it extends downward from $P$ Think of the vertical line through $P$ as a real axis, labelled with numbers going positive in the upward direction.

coffeemath
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What the picture shows is that the right triangle $OSQ$ is similar to the right triangle $OPR$. In particular, $$\tan(h)=\frac{\sin(h)}{\cos(h)}$$ by definition, and trivially, $$\tan(h)=\frac{\tan(h)}1.$$ The length of $OS$ is $\cos(h)$ ahd the length of $SQ$ is $\sin(h)$ by definition. Since it's a unit circle, then the length of $OP$ is $1$, so by similarity, the length of $PR$ is therefore $\tan(h)$.

Cameron Buie
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  • Exactly! The $\tan(t)$ that lies tangent to the circle isn't the function of the graph. It's just the length of the line. – Brian Silva Mar 25 '13 at 13:36
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Let us find the intersection of the Unit Circle is $x^2+y^2=1$ with the line $y=\tan t$

So, $x^2=1-y^2=1-\tan^2t\implies x^2+\tan^2t-1=0$

This is a Quadratic equation in $x$

For tangency, the two roots must be equal i.e., the discriminant $(0)^2-4\cdot1\cdot(\tan^2t-1)$ must be $0$

$\implies \tan^2t=1\implies \tan t=1$ as $0^\circ<t<90^\circ$

$\implies x=0, y=1$

So, $y=\tan t$ is tangent of the Unit Circle at $(0,1)$

lab bhattacharjee
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