Proof:
Let f(θ) = a cos θ + b sin θ
Also, let a = r sin α and b = r cos α.
Therefore we have
a2 + b2 = (r2 sin2 α) + (r2 cos2 α) and a/b = r sin α / r cos α
a2 + b2 = r2 (sin2 α + cos2 α) and a/b = tan α
r = ---- (i) and tan α = a/b ---- (ii)

Substituting the values of a and b in f(θ), we have

f(θ) = r sin α cos θ + r cos α sin θ
f(θ) = r(sin α cos θ + cos α sin θ)
f(θ) = r sin (α + θ) ---- (iii)
Knowing that
– 1 ≤ sin (α + θ) ≤ 1 for all θ
and multiplying throughout by r, we have
– r ≤ r sin (α + θ) ≤ r for all θ

Using (i) & (iii), we have

≤ f(θ) ≤ for all θ
≤ a cos θ + b sin θ ≤ for all θ
Hence, maximum and minimum values of a cos θ + b sin θ are and respectively.