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Prove that sin^2x+cos^2x=1 using derivatives.
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We'll apply s consequence of Lagrange's rule and we'll conclude that a function is a constant if and only if its derivative is cancelling.
We'll assign a function to the sum of trigonometric functions:
f(x) = (sin x)^2 + (cos x)^2
We'll calculate it's derivative using chain rule:
f'(x) = 2 sin x*(sin x)' + 2cos x*(cos x)'
f'(x) = 2sin x*cos x - 2cos x*sin x
We notice that the terms of the difference are the same, so we could eliminate them.
f'(x) = 0.
Since the derivative is zero, that means that f(x) is constant.
We'll prove that the constant has the value 1.
For x = 0=>(sin 0)^2 + (cos 0)^2 = 0 + 1 = 1
For x = pi/2 => (sin pi/2)^2 + (cos pi/2)^2 = 1 + 0 = 1
For x = pi => (sin pi)^2 + (cos pi)^2 = 0 + (-1)^2 = 0 + 1 = 1
For x = 2pi =>(sin 2pi)^2 + (cos 2pi)^2 = 0 + 1 = 1
The function f(x) = (sin x)^2 + (cos x)^2 is a constant and it has the value 1.
Posted by giorgiana1976 on February 21, 2011 at 1:59 AM (Answer #1)
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