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!

Similar Questions
  1. solve sin(theta)=-1 for all real values of (theta)I don't understand but by following the example in my book; sec(theta)=-1 from
    1. answers icon 3 answers
  2. Suppose that \Theta takes values in [0,1] and its PDF is of the formf_{\Theta }(\theta )= a\theta (1-\theta )^2,\ \ \
    1. answers icon 1 answer
  3. 1. Which expression is equivalent to cos 2theta for all values of theta ?cos^2 theta – sin^2 theta ~ cos^2 theta – 1 1 – 2
    1. answers icon 1 answer
  4. Solve sin theta=-1 for all real values of theta.Is it theta=3pi/2 +2pi n for n=0+ infinity?
    1. answers icon 2 answers
more similar questions