Thread: Where am I messing up? (Proof) [cos3x = 4cos^3x-3cosx]

1. Where am I messing up? (Proof) [cos3x = 4cos^3x-3cosx]

Hi, I am doing a proof and I have it all down but in the last step it seems as though I distributed my negatives incorrectly. However, I cannot seem to find where. Any help would be great! Thanks!

cos3x = 4cos^3x-3cosx is the proof. So I did:

cos3x =
cos (2x + x) =
cos2xcosx+sin2xsinx =
cosx (2cos^2x-1) + sinx(2sinxcosx) =
2cos^3x - cosx + 2sin^2xcosx =
2cos^3x - cosx + 2(1-cos^2x)cosx =
2cos^3x - cosx + 2cosx - 2cos^3x

However, 2cos^3x - cosx + 2cosx - 2cos^3x Does NOT = 4cos^3x-3cosx
Whereas 2cos^3x - cosx - 2cosx + 2cos^3x would

Thanks for any help!

2. Originally Posted by SMA777

cos (2x + x) =
cos2xcosx+sin2xsinx =
It's actually a $\displaystyle -$ sign instead $\displaystyle +.$ (Second step.)

3. Originally Posted by Krizalid
It's actually a $\displaystyle -$ sign instead $\displaystyle +.$ (Second step.)
Er... maybe I'm just being dense, but why?

Oh, I copied my Sum/Difference formula incorrectly! Got it. Thank you!

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y = 4sin3x-4cos3x ubahlah dalam bentuk k cos (x-A)

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