## 275- A problem from Homework 2

Quite a few of you had difficulties with problem 18 of the Additional and Advanced Exercises for Chapter 10, so I am posting a solution here.

The problem asks to derive the trigonometric identity $\sin(A-B)=\sin A\cos B-\cos A\sin B$ by forming the cross product of two appropriate vectors.

In problems that involve trigonometry or geometry, it is convenient to begin with vectors that have some clear geometric meaning related to the problem at hand, so it seems natural to consider the vectors $\vec u=(\cos A,\sin A,0)$ and $\vec v=(\cos B,\sin B,0)$.

These are two vectors in the plane, but we look at them as vectors in 3-D, as we should, since we want to look at their cross product, and this is only defined for vectors in 3-D.

So: $\vec u$ is a vector of size 1 ( $\|\vec u\|=\sqrt{\cos^2 A+\sin^2 A+0^2}=1$) that forms an angle of $A$ radians with the $x$-axis (measured counterclockwise). Similarly, $\vec v$ is a vector of size 1 that forms an angle of $B$ radians with the $x$-axis (measured counterclockwise).

Now: We need to analyze the angle between $\vec u$ and $\vec v$, which seems to be the technical point of this exercise, so let’s do this very carefully. This angle is the angle measured starting at $\vec u$ and moving counterclockwise until we find $\vec v$, is usually $B-A$, but it may be $2\pi-A+B$ if, for example, $\vec u$ and $\vec v$ are vectors in the first quadrant and $B. (Although we can “ignore'' this case since the sine function is periodic with period $2\pi$.)

Similarly: The direction of $\vec u\times\vec v$ is obtained by the Right-Hand Rule, meaning $\vec u\times \vec v$ is a vector perpendicular to the plane spanned by $\vec u$ and $\vec v$ (the $xy$-plane), but it may be a positive (or zero) multiple of ${\bf k}$, or a negative multiple of ${\bf k}$, depending on whether the angle between $\vec u$ and $\vec v$ is smaller than (or equal to) $\pi$, or larger than $\pi$.

The magnitude of $\vec u\times \vec v$ is $\|\vec u\|\|\vec v\|\sin\theta$, where $\theta$ is either the angle $\alpha$ between $\vec u$ and $\vec v$, or $2\pi-\alpha$, whichever is between $\null0$ and $\pi.$

Putting these two bits of information (about direction and magnitude) together, we find that $\vec u\times\vec v=(0,0,\sin(B-A))$ if $0\le B-A\le\pi$. If $B-A>\pi$, then $\vec u\times\vec v=(0,0,-\sin(2\pi+A-B)$ but $\sin(2\pi+\alpha)=\sin(\alpha)=-\sin(-\alpha)$ for any $\alpha$, so also in this case $\vec u\times\vec v=(0,0,\sin(B-A))$.

Finally, $\vec u\times\vec v$, component-wise, is found by computing the formal determinant $\left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ \cos A&\sin A&0\\ \cos B&\sin B&0\end{array}\right|=(0,0,\cos A\sin B-\sin A\cos B)$. Comparing this expression with the one above, we find the desired identity.

(Actually, we find it with the roles of $A$ and $B$ reversed, but this is of course irrelevant. And of course this deduction only works for angles between $\null0$ and $2\pi$, but the identity is true in all other cases as well, thanks to the periodicity properties of sine and cosine.)