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anonymous
 one year ago
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
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's simple try (ab)^2=a^2+b^22ab

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's great. Thanks keerthy.
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