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 Acos B-cos Asin 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 2pi-A+B if, for example, vec u and vec v are vectors in the first quadrant and B<A. (Although we can “ignore” this case since the sine function is periodic with period 2pi.) 

Similarly: The direction of vec utimesvec v is obtained by the Right-Hand Rule, meaning vec utimes 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 utimes vec v is |vec u||vec v|sintheta, where theta is either the angle alpha between vec u and vec v, or 2pi-alpha, whichever is between null0 and pi. 

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

Finally, vec utimesvec 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&0end{array}right|=(0,0,cos Asin B-sin Acos 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 2pi, but the identity is true in all other cases as well, thanks to the periodicity properties of sine and cosine.)

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