Posted by Paul on Monday, December 1, 2008 at 8:19pm.
have you come across the identity
cos 2x = 2cos^2 x - 1
we can just apply this here.
cos 2x = 2(3/5)^2 - 1
= -7/25
No I have not come across that before.
then probably
cos2x = cos^2 x - sin^2 x
(I replaced sin^2 x with 1-cos^2 x to get the one I used)
Your question fits right into the application of the double and half angle formulas, and without these your question becomes a mess.
Oh I get it now. thanks for your help.
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