# Prove that arcsinx + arccosx = pi/2 .

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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)

=> 1

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:

f'(x)=0,

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**