中四A.MATH題=.=

Let rsin(x - k) = sin x - 2 cos x
r sin x cos k - r cos x sin k = sin x - 2cos x
so r cos k = 1....................(1)
and r sin k = 2................(2)
(1)^2 + (2)^2 gives
r^2 = 1^2 + 2^2 = 5, therefore r = sqrt 5.
(2)/(1) gives tan k = 2
therefore k = arctan 2.
So (sin x - 2 cos x)^2 = [r sin (x - k)]^2
its max. value = r^2 = 5.
so the min. value of the question = 1/5.