Let g(x, y) = cos(x + 3y). Evaluate g(6, -2).

Solution:

Given that

g(x, y) = cos(x + 3y)

Then evaluate the function at g(6, -2)

g(6, -2) = cos(6 + 3(-2))

= cos(6 - 6)

= cos(0)

= 1

Similarly if g(x, y) = sin(x + 3y) then g(6, -2) will be:

g(6, -2) = sin(6 + 3(-2))

= sin(6 - 6)

= sin(0)

= 0

Let g(x, y) = cos(x + 3y). Evaluate g(6, -2).

Summary:

If g(x, y) = cos(x + 3y) then g(6, -2) = 1