# How do I find tangent on the unit circle?

I know that $\sin=y$ and that $\cos=x$ so how do I find tangent? I've seen some diagrams where the tangent line touches the unit circle at $(1,0)$, I'm wondering how this is derived?

EDIT: Explanations using similar triangles are most intuitive to me, if things could be explained in those terms I'd be grateful!

Thank you!

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The tangent is orthogonal to the line which goes from the center to the point on which you want the tangent. As this point is $$(\cos(x),\sin(x))$$ the tangent. will be $$(-\sin(x),\cos(x))\cdot t+(\cos(x),\sin(x))$$

You can easily visualize this using an CAS like Mathematica, taking as input

Manipulate[
ParametricPlot[
{
{Cos[Pi*t],Sin[Pi*t]},
{Cos[x], Sin[x]} + {-Sin[x], Cos[x]}*t
},
{t, -1, 1}],
{x, -Pi, Pi}]


gives you a slider where you can go through every angle. If you like I can make some pictures.

A picture

What i am essentially doing is drawing at first the circle, than i take a point on the circle which is $(\cos(x),\sin(x))$. From here I make a line in the direction of $(-\sin(x),\cos(x))$. This essentialy is a line of the form $a t + b$ but here $a$ and $b$ are vectors.

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Congrats for making sense of the question, +1. –  1015 Mar 24 '13 at 20:40
@DominicMichaelis I don't understand your answer, the tangent should have the form $y=at+b$ where $a$ and $b$ are real! –  Sami Ben Romdhane Mar 24 '13 at 20:48
that doesn't in every case (for example try to give the tangent of the point x=1 in that form. @SamiBenRomdhane –  Dominic Michaelis Mar 24 '13 at 20:52
Why is it that the tangent should be orthogonal? –  Assad Mar 24 '13 at 20:56
Also furthermore how did you derive (−sin(x),cos(x))⋅t+(cos(x),sin(x))? I'm sorry this is just a little confusing to me, a bit more explanation would help, thank you. –  Assad Mar 24 '13 at 20:57

Take a point $\,(\cos x_0\,,\,\sin x_0)\,$ on the unit circle, and assume $\,\cos x_0\cdot\sin x_0\neq 0\,$ (otherwise the question is almost trivial) , so the slope of the radius in the circle to this point is

$$\frac{\sin x_0}{\cos x_0}$$

Since the tangent line to the circle in the above point is perpendicular to the radius at that point, the tangent line's slope must have slope

$$-\frac{\cos x_0}{\sin x_0}$$

So now you have a point on the tangent line and its slope: calculate its formula.

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When $\cos(x_0) \cdot \sin(x_0)=0$ this is not trivial take $\cos(x_0)=1$ and try to give the tangent with your method –  Dominic Michaelis Mar 24 '13 at 20:53
No, when $\,\sin x_0=0\,$ then we're either on $\,(1,0)\,$ or on $\,(-1,0)\,$ and it's almost trivial the tangent lines there are the vertical lines $\,x=1\,$(resp., $\,x=-1\,$ ) . My answer does not cover these cases, that's why I assumed what I did. –  DonAntonio Mar 24 '13 at 20:56

Edit: Personally I prefer all the other answers, here's another way to think about it if you want more ideas

If $x^2 + y^2 = 1$, then $2x + 2y \frac{dy}{dx} = 0$

Re-arranging, $\frac{dy}{dx} = \frac{-x}{y}$. As you say, $\cos \theta = x$ and $\sin \theta = y$, giving you the slope of a tangent line as $- \cot \theta$. You can then use the equation: $$y - \sin \theta = - \cot \theta (x - \cos \theta)$$

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From the article $148,150$ of The elements of coordinate geometry, by Loney, the equation of the tangent of the circle $x^2+y^2=a^2$ at $(x_1,y_1)$ is
$$xx_1+yy_1=a^2$$
As you have already identified, any point on the circle can be $(a\cos\theta,a\sin\theta)$ where $0\le \theta<2\pi$
So, the equation of the tangent becomes $$xa\cos\theta+ya\sin\theta=a^2\implies x\cos\theta+y\sin\theta=a\text{ as }a\ne0$$
For the Unit Circle $a=1,$ So, the equation of the tangent becomes $x\cos\theta+y\sin\theta=1$