Find the maximum and minimum value of the function y-sin ^ 2 (x) -x / 2 on the interval [0, pi / 2]

Let’s find the derivative of the function:

y ‘= (sin ^ 2 (x) – x / 2)’ = 2sin (x) cos (x) – 1/2.

We equate it to zero:

2sin (x) cos (x) – 1/2 = 0.

Using the double argument formula for sine we get:

sin (2x) – 1/2 = 0;

sin (2x) = 1/2.

The roots of an equation of the form sin (x) = a are determined by the formula: x = arcsin (a) + – 2 * π * n, where n is a natural number.

2x = arcsin (1/2) + – 2 * π * n;

x = π / 12 + – π * n.

One point belongs to a given interval:

x0 = π / 12.

The value of the function at this point will be:

f (π / 12) = (sin (π / 12)) ^ 2 – π / 24.