# What is calculus approach in proof of trig fundamental identities? (exemplification)

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The example that follows shows how to use calculus to prove the basic identity of trigonometry.

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

You should attribute a function f(x) to the equation `sin^2 x + cos^2 x = 1` such that `f(x) = sin^2 x + cos^2 x = 1` .

Notice that the function is a constant. If so, its derivative cancells.

You need to differentiate the expression `sin^2 x + cos^2x` such that:

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

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

Notice the terms of f'(x) cancel each other such that: `f'(x) = 0` .

Hence, the function `f(x) = sin^2 x + cos^2 x` is constant.

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

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

**Hence, the basic trigonometric identity `sin^2 x + cos^2x = 1` is verified using the property of the derivative of a constant function.**

In a similar way, the following trig identity can also be proved using calculus.

*tan2x + 1 = sec2x*

*Let **f(x) = sec2x - tan2x = 1*

Differentiate both sides with respect to x. If the identity is correct then we will show that *f’(x) = 0 since the derivative of a constant is zero.*

*f’(x) = 2*tan x(sec2x) - 2*sec x(sec x*tan x)*

*f’(x) = 0*

*Therefore f(x) is a constant function.*

*If x = 0, f(0) = sec2(0)- *

*tan2(0) = 1 - 0 = 1*

*If x = pi**/4, f(pi/4) = sec2(pi/4)- tan2(pi/4) = 2 - 1 = 1*

Verifying that f(x) = 1, thus proving the identity:

* *

*tan2x + 1 = sec2x*