My question stems from the following graphic:

enter image description here

It aims to explain the following relationship:

$$\hat r = cos\ \theta \ \hat i + sin\ \theta \ \hat j$$ $$\hat \theta = -sin\ \theta \ \hat i + cos\ \theta \ \hat j $$

Now, judging from the graphic, $\hat r$ is derived from considering trigonometric properties of the triangle formed by hypotenuse $\hat r$ and its sine and cosine components as its legs. Since $\theta$ is at the tail of $\hat r$, its $x$ component is $cos \ \theta$ and its y component $sin \ \theta$. However, going by the logic of the derivation of $\hat r$, I'd expect the $x$ component of $\hat \theta$ to be $cos\ \theta_2$ and $y$ to be $sin\ \theta_2$. Note that I also added the subscript $2$ to differentiate it between the $theta$ used in finding the components of $\hat r$, but it highlights my two main confusions.

  • Why is theta taken to be at the head of $\hat \theta$ so that its $y$ and $x$ components are the way they are, and if that's not why $x$ is $sin\ \theta$ and $y$ is $cos\ \theta$, what's the reason?
  • Sine and cosine $\theta$ were written twice in each triangle in question in this graphic. Are they the same magnitude? And if not, why haven't they been distinguished by something like $\theta$ and $\theta_2$?

I feel like its highlighting a key trigonometric property I've forgotten, but since I don't know exactly what I'm not understanding I can't look up such a detail.


$\hat \theta$ is a unit vector which is at right angles to the unit vector $\hat r$.
If this is so then $\hat \theta \cdot \hat r =0$

So you need to find a vector $\hat \theta = a \hat x + b \hat j$ such that the conditions $a^2+b^2 =1$ and $\hat \theta \cdot \hat r =0$ are satisfied.
You will find that there are two unit vectors $\hat \theta$ which satisfy those conditions and by convention the vector which is pointing in the anti clockwise is the one which is used.

I have annotated your diagram to try and show you what is going on.

enter image description here

You have a unit vector $\hat r$ in green with components $(+\cos \theta, +\sin \theta)$ shown in violet.

The unit vector $\hat r$ can be rotated with the locus of the head of the vector shown as a dashed black line which is an arc of a circle of radius one.

As you want $\hat \theta$ shown in red to be at right angles to $\hat r$ it has components $(-\sin \theta, +\cos \theta)$ shown in blue.

Simple geometry can be used to show that all angles $\theta$ are the same because $\hat \theta$ is perpendicular to $\hat r$.

  • $\begingroup$ Thanks so much, this was really helpful. If I understand correctly, would it be correct to say that the $\theta$ used to derive the components of $\hat \theta$ is the angle next to the head of $\hat \theta$, since it's the only candidate that could be the same $\theta$ used in $\hat r$? $\endgroup$ – sangstar Sep 25 '17 at 17:37
  • $\begingroup$ @sangstar I have labelled that angle $\theta$. $\endgroup$ – Farcher Sep 25 '17 at 17:45
  • $\begingroup$ Right, and that's because the other possible angle candidate has to be $90 - \theta$ since its complimentary to the angle of $\hat r$? $\endgroup$ – sangstar Sep 25 '17 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.