prove the identities?

Question

4(cos^6x+ sin^6x) = 1+ 3cos^2(2x)
2cot(2x) = cotx- tanx

Answer 1

4(cos^6(x) + sin^6(x)) = 1 + 3cos^2(2x)

Choose the left hand side, since it is more complex.

LHS = 4(cos^6(x) + sin^6(x))

Factor as a sum of cubes.

LHS = 4[ (cos^2(x) + sin^2(x)) (cos^4(x) - 2cos^2(x)sin^2(x) +
sin^4(x))

Note that cos squared x plus sine squared x is equal to one.

LHS = 4[ (cos^4(x) - 2cos^2(x)sin^2(x) + sin^4(x)) ]

LHS = 4 [ cos^2(x) (cos^2(x) - sin^2(x)) + sin^4(x) ]
LHS = 4 [ cos^2(x) (cos2x) + sin^4(x)]
LHS = 4 [ cos^2(x) (cos2x) + [sin^2(x)]^2 ]
LHS = 4 [ cos^2(x) (cos2x) + [1 - cos^2(x)]^2]
LHS = 4 [ cos^2(x) (cos2x) + 1 - 2cos^2(x) + cos^4(x) ]
LHS = 4 [ cos^2(x) (cos2x) + cos^4(x) - 2cos^2(x) + 1 ]

I give up.