Use the Cauchy-Schwarz inequality to prove that for all real values of a, b, and θ

the question wants us to use the cauchy-schwarz inequality to prove that for all real values of a, b, and θ,

(a cosθ + bsinθ)^2 ≤ a^2 + b^2

so the Cauchy-Schwarz inequality is | < u,v> | ≤ ||u|| ||v||

I'm having a difficult time figuring out what is my u and my v.
Step by step explanations would be immensely helpful because I'd like to understand the problem.

Someone wrote this as an answer but I'm not sure what it means. Here it is:

with vectors (a,b) and (cosθ , sinθ) then
(a,b).(cosθ,sinθ) <=(a^2+b^2)(cos^2(θ)+sin^2(θ))
=> (acos(θ)+bsin(θ))^2<=a^2+b^2 since cos^2(θ)+sin^2(θ)=1

If anyone could explain the answer above or just explain the solution it would be much appreciated. Thank you so very much.

Note that
a cos θ + b sin θ = (a, b) · (cos θ, sin θ).

This suggests using u = (a, b) and v = (cos θ, sin θ).

Substituting this into the Cauchy-Schwarz Inequality yields
|(a, b) · (cos θ, sin θ)| ≤ ||(a, b)|| ||(cos θ, sin θ)||
==> |a cos θ + b sin θ| ≤ √(a^2 + b^2) √(cos^2(θ) + sin^2(θ))
==> |a cos θ + b sin θ| ≤ √(a^2 + b^2) * 1

Square both sides:
(a cos θ + b sin θ)^2 ≤ a^2 + b^2.

I hope this helps!