# Deriving trigonmetric identities, little confused?

• Oct 5th 2012, 09:29 PM
mathNoob89
Deriving trigonmetric identities, little confused?
I was reading through a text book on trigometric identities and gave the definition:
• cos(a) = sin(a + 90) // (1a)
• sin(a) = cos(a - 90) // (2a)

then it went to derive another two new identities from the above identities
• cos(a) = -sin(a-90) // (1b)
• sin(a) = -cos(a+90) // (2b)

What I don't understand is steps involved from deriving (1a) to (1b) and (2a) to (2b), any help would be appreciated (Nod)
• Oct 5th 2012, 10:07 PM
MaxJasper
Re: Deriving trigonmetric identities, little confused?
Draw a circle and show sin & cos axes on it and relevant angles...then all will be clear.
• Oct 5th 2012, 10:36 PM
mathNoob89
Re: Deriving trigonmetric identities, little confused?
Thank you for your reply, although specifically I wanted to see how you could derive the identities algebraically. That is, cos(a) = sin(a + 90) to cos(a) = -sin(a-90) and similalrly for the other identity. If you could show the steps, that would be great :)
• Oct 5th 2012, 10:42 PM
MaxJasper
Re: Deriving trigonmetric identities, little confused?
$\displaystyle \cos (\alpha -90)=\sin (90) \sin (\alpha )+\cos (90) \cos (\alpha )$