The key to this problem lies in the use of implicit conditions: square a of s in+square a of A+cos = 1.

Because Sina +cosA= root number 2,

So (sinA+cosA) =2 squared,

Extend the above formula and get

Square of sin a+square of a A+cos+2 * Sinacosa = 2,

Substituting the conditions of square of sin a+square of a +cos A= 1, we can get

1+2*sinAcosA=2，

So sinAcosA= 1/2.

Similarly,

The fourth power of sin a+the fourth power of cos a = the square of (the square of sin a+the square of cos a)-the square of sin a of -2 * A+cos.

= 1 -2*(sinAcosA) squared.

= 1-2*( 1/2) squared

= 1/2

Trigonometric function is the same, as long as the flexible use:

Square of sin a+square of a A+cos = 1.

sin2*x=2sinxcosx

cos 2 * x = 1-2 * sinx = 2 * cosx- 1

Then this problem can be solved!