How to simplify SiNx ^ 4 + cosx ^ 4, What about SiNx ^ 4-cosx ^ 4

How to simplify SiNx ^ 4 + cosx ^ 4, What about SiNx ^ 4-cosx ^ 4

sinx^4+cosx^4
=(sinx^2+cosx^2)^2-2sinx^2cos^2
=1-2*[2sinxcosx]^2/4
=1-sin2x^2/2

The maximum of function f (x) = 3 (SiNx) 2-2sin2x + 11 (cosx) 2
The answer is 2 radical 5 + 1

f(x)
=3((sinx)^2+(cosx)^2)-2sin2x+8(cosx)^2
=3-2sin2x+8(cosx)^2
=3-2sin2x+4(1+cos2x)
=7+2*5^0.5 *sin(2x+y)
Where siny = 2 / (5 ^ 0.5), cosy = - 1 / (5 ^ 0.5)
So the maximum value of F (x) is 7 + 2 * 5 ^ 0.5, when 2x + y = π / 2
My result is different from your "answer is 2 root 5 + 1", but the method is right

Why SiNx + (1 / cosx) = 1 / 2sin2x + 1? Did you miss dividing by cosx? Because cosx [SiNx + (1 / cosx)] = 1 / 2sin2x + 1

Yes, if this equation doesn't hold, it can be checked by substituting a simple number, such as 30 ° and 45 ° etc