Trig - Double Angles Questions,

[Macleef]

Trig - Double Angles Questions, Please Help!

please close, solved all my problems myself...

 

[rocomath]

Double Angle Identities are as follows:
sin2x = 2sinxcosx

cos2x = cos^2x - sin^2x
= 2sin^2x - 1
= 2cos^2x - 1

first off, your double-angle identities are wrong

\cos{2x}=\cos^{2}x-\sin^{2}x

\cos{2x}=2\cos^{2}x-1

\cos{2x}=1-2\sin^{2}x

 

[Macleef]

first off, your double-angle identities are wrong

\cos{2x}=\cos^{2}x-\sin^{2}x

\cos{2x}=2\cos^{2}x-1

\cos{2x}=1-2\sin^{2}x

what are you talking about? they're exactly the same

 

[rocomath]

what are you talking about? they're exactly the same

your signs are off, i haven't checked your work though