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 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 and .

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: is a vector of size 1 () that forms an angle of radians with the -axis (measured counterclockwise). Similarly, is a vector of size 1 that forms an angle of radians with the -axis (measured counterclockwise).

Now: We need to analyze the angle between and , 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 and moving counterclockwise until we find , is usually , but it may be if, for example, and are vectors in the first quadrant and . (Although we can “ignore'' this case since the sine function is periodic with period .)

Similarly: The direction of is obtained by the Right-Hand Rule, meaning is a vector perpendicular to the plane spanned by and (the -plane), but it may be a positive (or zero) multiple of , or a negative multiple of , depending on whether the angle between and is smaller than (or equal to) , or larger than .

The magnitude of is , where is either the angle between and , or , whichever is between and

Putting these two bits of information (about direction and magnitude) together, we find that if . If , then but for any , so also in this case .

Finally, , component-wise, is found by computing the formal determinant . Comparing this expression with the one above, we find the desired identity.

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

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