cos (x-A) = a

cos (x-A) = a
sin (x-B) = b
prove cos^2 (A-B) = a^2 + b^2 - 2ab sin (A-B)


Solution:



M1

Letting p = x - A , and q = x - B ,
the statement in question becomes
cos^2 (q - p) = a^2 + b^2 - 2ab sin (q - p)

LHS
= (cos p cos q + sin p sin q)^2
= a^2 cos^2 q + b^2 sin ^2 p + 2ab sin p cos q

Note that the 1st term = a^2 (1 - b^2)
and the 2nd term = b^2 (1 - a^2) .
So, LHS becomes a^2 + b^2 - 2abH ,
where H = ab - sin p cos q = sin (q - p) .
QED.



M2

Basement cat's solution (see comments).
Besides the brute force, another apporach:

Let A, B be fixed.
Consider a, b to be a function of x .
f(x) = a^2 + b^2 - 2ab sin (A - B)
f'(x) = 2aa' + 2bb' - 2(ab' + a'b) sin (A - B)

And note that
sin (A - B) = sin (A - x + x - B) = a'b' + ab
(a')^2 + a^2 = 1
(b')^2 + b^2 = 1

Thus, f'(x) = 0 .
i.e. f(x) is a constant. In particular,
f(B) = cos^2 (A - B)