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

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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.**