Expand cos4A?

cos(4A)

= cos 2(2A)

= cos²(2A) - sin²(2A)

= (cos²A - sin²A)² - (2sinA cosA)²

= [cos⁴(A) - 2 sin²(A) cos²(A) + sin⁴(A)] - 4 sin²(A) cos²(A)

= cos⁴(A) - 6 sin²(A) cos²(A) + sin⁴(A)

Not clear whether you want a Mclaurin expansion.
cos 4A = cos^2 2A - sin^2 2A
= 1 - 2 sin^2 A - ( 2 sin A cos A)^2
= 1- 2 sin^2 2A - 4 sin^2 A cos^2 A
= 1 - 2 (2 sin A cos A)^2 - 4 sin^2 A cos^2 A
= 1- 4 sin^2 A cos^2 A - 4 sin^2 A cos^2 A
= 1- 8 sin^2 A cos^2 A

Well,

first of all, welcome to this forum from (maybe the only) french guy in the zone !! ;-)

I assume it is a LINEARISATION exercise : obtaining a polynomial P(cosx) of the vairiable (cosx)

cos(4x) = cos(2x + 2x) = cos(2x)cos(2x) - sin(2x)sin(2x)
= (2cos^2 - 1)^2 - sin(2x)^2
= (2cos^2 - 1)^2 - 4sin^2xcos^2x
= (2cos^2 - 1)^2 - 4cos^2x (1 - cos^2x)
= 4 cos^4x - 4cos^2x + 1 - 4cos^2x + 4cos^4x
= 8cos^4x - 8cos^2x + 1

cos(4x) = 8cos^4x - 8cos^2x + 1

et voilà !!

hope it' ll help !!

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to me, or anybody else, or course provided the answer deserves one !
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