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We have to prove that arc sin x + arc cos x = pi/2
The left hand side is
arc sin x + arc cos x
take the sine of the angle
sin (arc sin x + arc cos x)
=> sin (arc sin x)* cos (arc cos x) + cos (arc sin x)* sin (arc cos x)
use the relation sin (arc cos x) = cos (arc sin x) = sqrt (1 - x^2)
=> x*x + [sqrt (1 - x^2)]^2
=> x^2 + 1 - x^2
Take the sine of the right hand side
sin pi/2 = 1
We get the same value in both the cases
This proves that arc sin x + arc cos x = pi/2
By definition, a function is constant if and only if it's first derivative is cancelling.
We'll assign a function f(x) to the given expression arcsin x + arccos x.
f(x) = arcsin x + arccos x
We'll have to do the first derivative test.
f'(x) = (arcsin x + arccos x)'
f'(x) = [1/sqrt(1-x^2)] - [1/sqrt(1-x^2)]
We'll eliminate like terms:
Since the first derivative was zero, f(x) = constant.
To prove that the constant is pi/2, we'll put x = 1:
f(1) = arcsin 1 + arccos 1 = pi/2 + 0 = pi/2
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