You can put this solution on YOUR website! To avoid confusion, let's call the angle t
sin(t) and cos(t) are defined by a point on the unit circle, (x,y) as sin(t) = y, and cos(t) = x
Using a trig identity, we can write cos(t) as sqrt(1-sin^2(t))
So f = sin(t) + cos(t) = sin(t) + sqrt(1-sin^2(t)) = y + sqrt(1-y^2)
Since the hypotenuse is equal to 1, x and y are less than or equal to 1
The maximum value is obtained when df/dy = 0 -> 1 - y/sqrt(1-y^2) = 0
Solving for y, we get y =1/sqrt(2)
So the maximum value is y + sqrt(1-y^2) = 1/sqrt(2) + 1/sqrt(2) = 2/sqrt(2) = sqrt(2)
And the minimum value is obtained for y = 0 or y = 1, and the expression takes on the value of 1.
So we have shown that 1<=sin(t)+cos(t)<=sqrt(2)
