Question

cos^6A-sin^6A=cos2A(1-1÷4sin^2 2A) Prove

Answer 1

Factor the LS as a difference of cubes

LS = (cos^2 A - sin^2 A)(cos^4 A + (sin^2 A)(cos^2 A) + sin^4 A)
= cos (2A) (cos^4 A + sin^4 A + 2(sin^2 A)(cos^2 A) - (sin^2 A)(cos^2 A) )
= cos (2A) ( (cos^2 A + sin^2 A)^2 - sin^2 A cos^2 A )
= cos(2A) ( 1 - sin^2 A cos^2 A )
getting close ...

aside:
sin^2 A cos^2 A
= (sinAcosA)^2 , and since (sin 2A = 2sinAcosA)
= ( (1/2)sin (2A) )^2
= (1/4) sin^2 A

so LS = cos 2A)(1 - (1/4)sin^2 A)
= RS

Answer 2

Cot3A=Cot^3A_3CotA/3Cot^2A_1