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!

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

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

We can follow the steps below:

Step 1: Start by expanding the square expression on the left side of the inequality:

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

Step 2: Rearrange the terms by grouping the cosine and sine terms together:

(a^2 * cos^2 t + b^2 * sin^2 t) + 2ab * cos t * sin t

Step 3: Notice that the terms a^2 * cos^2 t and b^2 * sin^2 t can be rewritten using the trigonometric identity: cos^2 t + sin^2 t = 1:

(a^2 * (1 - sin^2 t) + b^2 * sin^2 t) + 2ab * cos t * sin t

Step 4: Simplify the expression:

(a^2 - a^2 * sin^2 t + b^2 * sin^2 t) + 2ab * cos t * sin t

Step 5: Combine like terms:

(a^2 + b^2 + (b^2 - a^2) * sin^2 t) + 2ab * cos t * sin t

Step 6: Observe that (b^2 - a^2) is a constant term and sin^2 t, cos t, and sin t are trigonometric functions that take values between -1 and 1. Therefore, (b^2 - a^2) * sin^2 t and 2ab * cos t * sin t are both less than or equal to zero.

Step 7: Using the inequality property that adding a non-positive value to a positive value does not change the inequality, we can conclude:

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

Therefore, we have proven that for all real values of a, b, and t (theta):

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