Solving Trig Identity Problem: Asin^2(wt) + Bcos^2(wt) = A = B

[DiracPool]

I can't quite work out this derivation I ran into which is essentially...Asin^2(wt) + Bcos^2(wt) = A = B. Is this correct?

I know that sin^2(wt) + cos^2(wt) = 1, but I can't reason out how the factoring works here? Any help?

 

[Char. Limit]

Where exactly did you run into it? Can you paste some context for us to look at?

Looking at what you wrote, as long as A = B, the full statement is true... but if that's the case, then you might as well just use A.

 

[DiracPool]

Where exactly did you run into it? Can you paste some context for us to look at?

Looking at what you wrote, as long as A = B, the full statement is true... but if that's the case, then you might as well just use A.

Actually, you can can find the full equation here at the bottom of the screen at 1:20 -

So A and B are not the same. In effect, its Asin^2(wt) + Bcos(wt) = constant, where the constant is Energy in this case, but the guy doesn't do the deriviation, he just presents the equation and I'm not clear on how he factored it.

 

[HallsofIvy]

I think you are misunderstanding what he is saying. It is not that "A sin^2(x)+ B cos^2(x)" is a constant- it is only for these specific, and related, A and B. And he is not using any trig identity. He is simply stating that the "A sin^2(x)+ B cos^2(x)" is the total energy and, because of "conservation of energy", must be constant. Of course, taking x= 0 gives "B" and taking x= pi/2 gives "A" so A and B must be equal to that constant and so to each other.