# [Quick Question]Rewriting Pythagorean Identity

• April 12th 2010, 05:11 PM
florx
[Quick Question]Rewriting Pythagorean Identity
http://i44.tinypic.com/35lavps.jpg

I need help with this problem. Please show your work and explain how you rewrote the Pythagorean equation into what they are asking. Thank you so much.
• April 12th 2010, 05:16 PM
pickslides
Quote:

Originally Posted by florx
http://i44.tinypic.com/35lavps.jpg

I need help with this problem. Please show your work and explain how you rewrote the Pythagorean equation into what they are asking. Thank you so much.

Remember $\sin ^2t+\cos^2t = 1$

$
2\cos^2t-1 = 2\cos^2t-(\sin ^2t+\cos^2t)= 2\cos^2t-\sin ^2t-\cos^2t = \cos^2t-\sin ^2t
$
• April 12th 2010, 05:37 PM
florx
Thank you so much for your quick answer for a). I had forgot you could make 1 = sin t + cos t

Anyways can you check to see if I did b) correctly?

b) cos 2t = (cos t)^2 - (sin t)^2
cos 2t = -sin^2t + cos^2t
cos^2t -sin^2t
(1-sin^2t) + (cos^2t -1)
cos 2t = -sin^2t + cos^2t
• April 12th 2010, 06:26 PM
harish21
Quote:

Originally Posted by florx
Thank you so much for your quick answer for a). I had forgot you could make 1 = sin t + cos t

Anyways can you check to see if I did b) correctly?

b) cos 2t = (cos t)^2 - (sin t)^2
cos 2t = -sin^2t + cos^2t
cos^2t -sin^2t
(1-sin^2t) + (cos^2t -1)
cos 2t = -sin^2t + cos^2t

If you look at pickslides' post, you can see that he has already done part(b) for you.

$2\cos^2t-1 = 2\cos^2t-(\sin ^2t+\cos^2t)= 2\cos^2t-\sin ^2t-\cos^2t = \cos^2t-\sin ^2t$

Do you know that $2\cos^2t-1 = \cos2t$? In fact, this is the part(a) of your question!

And your part (b) is thus proved!
• April 12th 2010, 06:50 PM
florx
Oh thanks I didn't notice that.