verify the idenity? (acost-bsint)^2+(asint+bcost)^2=a^2+b^2

Question

zehanz · Answer

First, you have to know how to expand forms like (a+b)² and (a-b)².
Can you do that?

anonymous · Answer

(acost + bsint) (acost-bsint) for the (a-b)^2?

zehanz · Answer

No, I mean (acost-bsint)² expands just like (p-q)², where p=acost and q=bsint,
So exapnd it first, then handle the other one.

zehanz · Answer

Let me help you: (p-q)²=p²-2pq+q², so (acost-bsint)²=...

anonymous · Answer

(acost-bsint)^2= (acost^2-2(acost)(bsint)+bsint^2?

zehanz · Answer

It is $a^2\cos^2t-2ab\sin t \cos t + b^2 \sin^2 t$

Then, (asint+bcost)² expands to:

 $a^2\sin^2 t+2ab\sin t \cos t +b^2 \cos^2 t $

zehanz · Answer

What happens if you add these two beasts?

anonymous · Answer

turns into the right side? the middle section cancels, right?

zehanz · Answer

Yes, they cancel, so we are left with:

$a^2(\cos^2t+\sin^2t)+b^2(\sin^2t+\cos^2t)$

What happens with the stuff between the brackets?

anonymous · Answer

turns into 1, due to the Pythagorean identity. You made it look so easy, hopefully it sticks in my head.

zehanz · Answer

OK, next time you will know what to do!

anonymous · Answer

Thank you, much appreciated!