Problem: Find the value of f((2pi)/3) given f(0) = cos20 cose + sine sine cose - sine sin^2 theta -cose -sino cose 0 ff(0) =.

Solution:

We are given the value of f(0) which is:

f(0) = cos20 cose + sine sine cose - sine sin^2 theta -cose -sino cose 0 ff(0) =

We can simplify this expression using trigonometric identities:

f(0) = cos20 + sin20 - sin^2(theta) - cos20 - cos20 = -cos20 - sin20 - sin^2(theta)

Now we can use this expression to find the value of f((2pi)/3). First, we need to find the angle (2pi)/3 in terms of cos and sin:

cos((2pi)/3) = -1/2
sin((2pi)/3) = sqrt(3)/2

Now we can use these values to find f((2pi)/3):

f((2pi)/3) = cos20 cos((2pi)/3) + sin20 sin((2pi)/3) - sin((2pi)/3)^2 - cos20 - sin20

= cos20*(-1/2) + sin20*(sqrt(3)/2) - (sqrt(3)/2)^2 - cos20 - sin20

= -cos20 - sin20 - 3/4

Therefore, the value of f((2pi)/3) is -cos20 - sin20 - 3/4.

Conclusion:

- We were given the value of f(0) and we used trigonometric identities to simplify it.
- We found the values of cos((2pi)/3) and sin((2pi)/3) and used them to find f((2pi)/3).
- The final value of f((2pi)/3) is -cos20 - sin20 - 3/4.