Simplify the expression:
√[sin4x+4cos2x] - √[cos4x+4sin2x]
Source: AMC12 2002
We have:
sqrt[(sin x)^4+4*(cos x)^2] - sqrt[(cos x)^4+4*(sin x)^2]
Lets start with utilizing the identity (sin x)^2 + (cos x)^2 = 1 to express each radicand in terms of just sin or just cos:
= sqrt[(sin x)^4+4-4*(sin x)^2] - sqrt[(cos x)^4+4-4*(cos x)^2]
Complete the squares in each radicand:
= sqrt[((sin x)^2-2)^2] - sqrt[((cos x)^2-2)^2]
At this point the radicals can be canceled with the squares, but pay attention to what the principal values of the roots are; 2-(sin x)^2 and 2-(cos x)^2 are positive. (This is the "trap" Jer alluded to in the comments) Thus we get:
= (2-(sin x)^2) - (2-(cos x)^2)
Now just a final bit of simplification:
= (cos x)^2 - (sin x)^2 = cos(2x).
Solution
