You also may use the following alternative method, hence, considering the function `f(x) = sin^2 x + cos^2 x = 1` . If differentiating the function f(x) with respect to x you arrive at the result f'(x) = 0, hence, the function f(x) is a constant. All you need to do is to determine if the value of the constant is 1.

Differentiating with respect to x yields:

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

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

`f'(x) = 0`

The first step proves that f(x) is a constant function. You need to determine the constant such that:

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

`x = pi/2 => f(pi/2) = sin^2 (pi/2) + cos^2(pi/2) = 1 + 0 = 1`

`x = pi => f(pi) = sin^2pi + cos^ 2pi = 0 + (-1)^2 = 1`

`x = 3pi/2 => f(3pi/2) = sin^2 (3pi/2) + cos^2(3pi/2) = (-1)^2 + 0 = 1`

`x = 2pi => f(2pi) = sin^22pi + cos^ 2 2pi= 0 + 1 = 1`

**Hence, using derivatives yields that for all values of x `sin^2 x + cos^2 x = 1.` **

In a right triangle, according to the Pythagorean theorem, we have:

a ² + b ² = c ².

Thus, if x is a measure of the angle B, then:

sin ² x + cos ² x = (b / a) ² + (c / a) ² =

b ² / a ² + c ² / a ² =

(b ² + c ²) / a ² =

Indeed,

a ² / a ² = 1.

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You can draw a picture to show. It is even more intuitive.