Question

Prove that for all real values of a, b, t (theta):

(a * cos t + b * sin t)^2 <= a^2 + b^2

I will be happy to critique your work. Start on the left, square it,

(a * cos t + b * sin t)^2 =
a^2 (1 - sin^2t) + 2ab sin t cost+ b^2 (1 - cos^2 t)=
a^2 + b^2 - (a sin t - b cos t)^2
<= (a^2 + b^2)
because (a sin t - b cos t)^2 mus be positive or zero.

I don't know how you figured that out, but thanks. That works perfectly!