How to prove that sin^2 x+cos ^2 x=1 using derivatives?

We'll assign a function to the Pythagorean expression (sin
x)^2 + (cos x)^2 = f(x).

From Pythagorean identity, we
notice that f(x) = 1.

We also know that the derivative of a
constant function is cancelling.

Therefore, if we'll
differentiate f(x) with respect to x, the result has to be
zero.

Let's check if it's
true.

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

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

We'll eliminate like terms and we'll
get:

f'(x) = 0

This result
emphasize the fact that the function is a constant.

Let's
see if the sum of the squares of the sine and cosine functions of the same angle is
1.

We'll put x = 0.

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

We'll put x =
pi/2

(sin pi/2)^2 + (cos pi/2)^2 = 1 + 0 =
1

We'll put x = pi

(sin pi)^2
+ (cos pi)^2 = 0 + (-1)^2 = 1

We'll put x =
2pi

(sin 2pi)^2 + (cos 2pi)^2 = 0 + 1 =
1

We notice that the Pythagorean identity,
(sin x)^2 + (cos x)^2 = 1, is verified for any value of
x.