Prove that arcsinx + arccosx = pi/2 .
Take a right triangle with the hypotenuse equal to 1.
Now let one of the other sides be equal to x. If the third side is s, s^2 + x^2 = 1
=> s^2 = 1 - x^2
=> s = sqrt (1- x^2)
Now for the triangle, if the angles are a and b.
sin(a) = x
cos(a) = sqrt(1 - x^2)
sin(b) = sqrt(1 - x^2)
cos(b) = x
Now, sin(a + b)
=> sin(a)cos(b) + cos(b)sin(a)
=> x^2 + (1-x^2)
Also, sin (pi/2) = 1.
So we have a + b = pi/2
But a = arc sin (x) and b = arc cos (x).
Therefore arc sin (x) + arc cos (x) = pi/2
We'll assign a function f(x) to the given expression arcsin x + arccos x.
f(x) = arcsin x + arccos x
According to the rule, a function is constant if and only if it's first derivative is cancelling. 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