I have this identity: cos24α=cos22α−sin⁡6αsin⁡2α If I write this like as: (cos⁡4α+cos⁡2α)⋅(cos⁡4α−cos⁡2α)=−sin⁡6αsin⁡2α I can use Werner and prostaferesis formulas and I find the identity (1). But if we suppose of not to use these formulas I have done a try writing: cos24α=cos22α−sin⁡6αsin⁡2α (1) [cos⁡(2(2α))]2=(2cos⁡2α−1)2−sin⁡6αsin⁡2α (2) cos42α−2sin22αcos22α+sin42α=(2cos2α−1)2−sin⁡6αsin⁡2α (3) Then I have abandoned because I should do long computations and I think that is not the right way.

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I have this identity:  cos24α=cos22α−sin⁡6αsin⁡2α  If I write this like as:  (cos⁡4α+cos⁡2α)⋅(cos⁡4α−cos⁡2α)=−sin⁡6αsin⁡2α  I can use Werner and prostaferesis formulas and I find the identity (1).  But if we suppose of not to use these formulas I have done a try writing:  cos24α=cos22α−sin⁡6αsin⁡2α             (1)  [cos⁡(2(2α))]2=(2cos⁡2α−1)2−sin⁡6αsin⁡2α    (2)  cos42α−2sin22αcos22α+sin42α=(2cos2α−1)2−sin⁡6αsin⁡2α          (3)  Then I have abandoned because I should do long computations and I think that is not the right way.

Donald Cheek · Accepted Answer

I think that we can establish the identity without too much work; it certainly isnt

lalilulelo2k3eq · Accepted Answer

OK, I think I have something a little more straightforward. If I am using some of the forbidden fruit, I apologize. First of all, as hamam already noted, I&#039;m going to let 2α=x to save my sanity. So I want to show thatcos22x=cos2x−sin⁡3xsin⁡xAside from the double-angle formula for cos⁡2x, I will need the triple angle formulacos⁡3x=cos⁡2xcos⁡x−sin⁡2xsin⁡x=(2cos2x−1)cos⁡x−2sin2xcos⁡x=2cos3x−cos⁡x−2(1−cos2x)cos⁡x=4cos3x−3cos⁡xNote that cos⁡2x=cos⁡(3x−x)=cos⁡3xcos⁡x+sin⁡3xsin⁡x, so the original formula is equivalent tocos2x−cos22x+cos⁡3xcos⁡x=cos⁡2x    (*)Well, by the triple angle formula above,cos⁡2x+cos⁡22x−cos⁡2x=(2cos2x−1)+(2cos2x−1)2−cos2x=4cos4x−3cos2x=cos⁡3xcos⁡xand this establishes (*)

Vasquez · Accepted Answer

hint Put a=2α and prove simply that cos2⁡(2a)=cos2⁡(a)−sin⁡(3a)sin⁡(a) using 2sin⁡(3a)sin⁡(a)= cos⁡(3a−a)−cos⁡(3a+a)

I have this identity: \cos^2 4\alpha=\cos^2 2\alpha−\

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