32sin4acos2a=cos6a-2cos4a-cos2a+2prove that

cos6a = cos(4a+2a) = cos4a cos2a - sin4a sin2a

cos4a = 2cos^2(2a)-1
see what you can do with that

To prove the given equation 32sin^4(a)cos^2(a) = cos^6(a) - 2cos^4(a) - cos^2(a) + 2, we will simplify both sides of the equation step by step.

Starting with the left-hand side (LHS):
LHS = 32sin^4(a)cos^2(a)

We know that sin^2(a) = 1 - cos^2(a). Using this identity, we can rewrite sin^4(a):
sin^4(a) = (sin^2(a))^2 = (1 - cos^2(a))^2

Now, substituting this into the LHS, we get:
LHS = 32(1 - cos^2(a))^2cos^2(a)

Expanding the square:
LHS = 32(1 - 2cos^2(a) + cos^4(a))cos^2(a)

Simplifying further:
LHS = 32cos^2(a) - 64cos^4(a) + 32cos^6(a)

Now, let's simplify the right-hand side (RHS):
RHS = cos^6(a) - 2cos^4(a) - cos^2(a) + 2

Combining like terms:
RHS = 32cos^6(a) - 64cos^4(a) + 32cos^2(a)

As we can see, after simplification, the LHS and RHS are identical expressions. Therefore, we have proven that 32sin^4(a)cos^2(a) = cos^6(a) - 2cos^4(a) - cos^2(a) + 2.