三角恆等式

I read the following from 船山筆記：

cos(x-A) = a
sin(x-B) = b
prove cos²(A-B) ≡ a² + b² - 2ab sin(A-B)

Two readers of the blog gave their solutions. One directly expanded cos²(A-B) ≡ cos²((x-B)-(x-A)) using the sum-of-angle formula. The other one cleverly noted that the LHS of the identity does not depend on x. So he/she differentiated the RHS and found that the derivative is indeed zero. Hence RHS is a constant function of x and the identity can be proved by putting x=B.

The two readers and the blogger, though, did not explain how the above trigonometric identity arises. So I did some calculations and found a way to explain the identity in a more natural manner. Here we go:

Replace B by B-π/2. Then the above problem becomes:

Let
cos(A-x) = a,
cos(B-x) = b.
Show that sin²(A-B) ≡ a² + b² - 2ab cos(A-B) ............ (#)

One may immediately recognise that the RHS of (#) is the squared magnitude of the difference of two complex numbers:

a² + b² - 2ab cos(A-B)
= | ae^iB - be^iA |²
= | (e^i(A-x) - e^-i(A-x)) e^iB/2 - (e^i(B-x) - e^-i(B-x)) e^iA/2 |²
= | (e^i(A+B-x) - e^i(-A+B+x) - e^i(A+B-x) + e^i(A-B+x))/2 |²
= | (-e^i(-A+B+x) + e^i(A-B+x))/2 |²
= | e^ix (e^i(A-B) -e^-i(A-B))/2 |²
= | e^ix i sin(A-B) |²
= sin²(A-B)

Now one can see why the RHS of (#) is independent of x.

The above result can be used to prove another trigonometric identity in 船山筆記, namely,

sin² B ≡ cos² A + cos²(A+B) - 2 cos A cos B cos(A+B).

Proof. Put x=0 and replace B by A+B in (#), we are done.