Hi,
In lecture 7 it is stated that,

Sqaureroot [ ( 1 - cost)^2 + sin^2t ]

is equal to

Sqaureroot [ 1- 2cost + cos^2t + sin^2t ]

I don't understand where the 2cost in the second expression came from. I've tried playing around with some trig identifies with no luck.

Could someone please explain it?
Thanks.

Question

Hi, 
In lecture 7 it is stated that,

Sqaureroot [ ( 1 - cost)^2 + sin^2t ]

is equal to

Sqaureroot [ 1- 2cost + cos^2t + sin^2t ]

I don't understand where the 2cost in the second expression came from. I've tried playing around with some trig identifies with no luck.

Could someone please explain it? 
Thanks.

anonymous · Answer

it's simple  try  (a-b)^2=a^2+b^2-2ab

anonymous · Answer

That's great. Thanks keerthy.