1. ## Inverse Functions

I'm not quite sure how to tackle this question. It asks;

a) Find the value of x such that $\displaystyle sin^{-1}x = cos^{-1}x$.

b) On the same diagram, sketch the graphs of $\displaystyle y = sin^{-1}x$, $\displaystyle y = cos^{-1}x$ and $\displaystyle y = sin^{-1}x + cos^{-1}x$.

With this second part, I'm able to sketch the graphs of $\displaystyle y = sin^{-1}x$ and $\displaystyle y = cos^{-1}x$, but I'm not sure how to sketch the third...

2. Originally Posted by Flay
I'm not quite sure how to tackle this question. It asks;

a) Find the value of x such that $\displaystyle sin^{-1}x = cos^{-1}x$.

b) On the same diagram, sketch the graphs of $\displaystyle y = sin^{-1}x$, $\displaystyle y = cos^{-1}x$ and $\displaystyle y = sin^{-1}x + cos^{-1}x$.

With this second part, I'm able to sketch the graphs of $\displaystyle y = sin^{-1}x$ and $\displaystyle y = cos^{-1}x$, but I'm not sure how to sketch the third...
a) $\displaystyle x = \frac{\pi}{4}$:

Let $\displaystyle \cos^{-1} x = \alpha \Rightarrow x = \cos \alpha$.

Then $\displaystyle \sin^{-1} x = \alpha \Rightarrow x = \sin \alpha \Rightarrow \cos \alpha = \sqrt{1 - x^2}$.

Therefore $\displaystyle x = \sqrt{1 - x^2} \Rightarrow x^2 = 1 - x^2 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$.

Discard extraneous solutions: $\displaystyle x = \frac{\pi}{4}$.

This solution should be obvious to you in hindsight (if not foresight).

b) Use addition of ordinates to sketch $\displaystyle y = \sin^{-1}x + \cos^{-1}x$. You will find that $\displaystyle y = \frac{\pi}{2}$ and this result ought not surprise you in hindsight.