Step-by-step explanation:
To prove that \[x^2 + y^2 = 1\] given \[\cos^{-1}(x) + \cos^{-1}(y) = \frac{\pi}{2}\], we will use the fact that the cosine function is restricted to the range \([-1, 1]\).
We start by examining the relationship between the arccosine function and the cosine function. The arccosine function \(\cos^{-1}(x)\) gives the angle \(\theta\) such that \(\cos(\theta) = x\), where \(\theta\) is restricted to the interval \([0, \pi]\).
Now, let's consider the equation \[\cos^{-1}(x) + \cos^{-1}(y) = \frac{\pi}{2}\]. This equation implies that the sum of two angles whose cosines are \(x\) and \(y\) is \(\frac{\pi}{2}\). Since the cosine of an angle in the interval \([0, \pi]\) is positive in the first and second quadrants, this implies that both \(x\) and \(y\) must be positive.
Now, let's consider the right-angled triangle with sides of length \(x\), \(y\), and the hypotenuse 1. By the Pythagorean theorem, we have \[x^2 + y^2 = 1\].
Therefore, we have proved that \[x^2 + y^2 = 1\] given \[\cos^{-1}(x) + \cos^{-1}(y) = \frac{\pi}{2}\].
