# Thread: [Quick Question]Rewriting Pythagorean Identity

1. ## [Quick Question]Rewriting Pythagorean Identity

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.

2. Originally Posted by florx

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

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

3. 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

4. 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.

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

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

And your part (b) is thus proved!

5. Oh thanks I didn't notice that.