# What is the value of x if cos^-1 x = 90-sin^-1 x?

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Consider a right angled triangle with hypotenuse a and sides ax and a*sq rt(1-x^2), where x <= 1.

Then the angles of the triangle opposite to a is 90 degree, opposite to ax is sin ^-x and opposite to the side a*sqrt(1-x^2) is cos^-1x.

Therefore cos^-1x+sin^-1 = 90 degree (ot pi/2 radians is an identity.

The value of sin and cos ratios are in the interval {-1, +1}.

So cos^-1x+sin^-1x= 90 deg or cos^-1x = 90 - sin ^-1x is true **for all x for which -1 < = x < = +1**.

We'll re-write the identity, using arcfunctions:

arccos x = pi/2 - arcsin x

We'll shift arcsin x to the left side and we'll get:

arccos x + arcsin x = pi/2

It is obvious that for x = 1, we'll get:

arcsin 1 + arccos 1 = pi/2 + 0 = pi/2

We have to prove that x = 1 is the only solution.

We'll assign a function f(x) to the given expression arcsin x + arccos x.

f(x) = arcsin x + arccos x

By definition, 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

**A continuous differentiable function is also bijective, therefore it is injective, so x = 1 is the only solution for the equation arccos x = pi/2 - arcsin x.**