let f(x)= sin^2 x + cos^2 x

Since we need to prove that f(x)=1

Then the first dericative must be zero.

f'(x)= 2cosx sinx-2sinxcos x=0

now we know that f(x) is a constant number (C)

==> f(x)= C

==> but f(0) = sin^2 0 + cos^2 0 = 1

Then f(x)=1

THen sin^2 x + cos ^2 x =1

To prove that (sinx)^2+(cosx)^2 = f(x)

Differentiating, we get:

2sinxcosx +2cosx (-sinx) = f'(x).

So = f'(x) .

So f'(x) = k a constant.

Bur at x= pi/2, sinx =1 and cosx =0. So (sinx)^2+(cosx)^2 = 1+0 = 1 is the LHS value.

But RHS value k. Therefore k =1. Or

(sinx)^2 + (cosx)^2 = k = 1.

If we consider the expression (sin x)^2 + (cos x)^2 as a function f(x), in order to demonstrate that f(x) = 1, so is a constant, we have to calculate the first derivative of this function f(x). If this derivative is 0, that means that f(x) = 1, is a constant function, knowing the fact that a derivative of a constant function is 0.

f'(x) = [(sin x)^2 + (cos x)^2]'

f'(x) = 2sin x*cos x + 2cos x*(-sinx)

f'(x) = 2sin x*cos x - 2sin x*cos x = 0, so that means that f(x) = ct.

But f(0) = (sin 0)^2 + (cos 0)^2 = 0 + 1 = 1