Check out some similar questions!

amstepp · Mar 2, 2011, 09:37 PM

we had this extra credit problem the hint that came with it was:
the usual method of proving identities does not work in this case, since we do not have any identities to use for inverse trigonometric functions. So, you will have to figure out a completely different line of attack. You need to somehow use the fact that

Cos theta = sin( pi/2 - theta). One way would be to sart with, "Let A=sin^-1x and B= cos^-1x," and proceed from there.

Unknown008 · Mar 3, 2011, 07:31 AM

Is that it?

$\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}$

Are you allowed yo use a graphical approach?

Code:

          A
         /|
        / |
       /  |
      /   |
     /    |
    /     |
   /      |
B /_______| C

$\cos^{-1} $\frac{BC}{AB}$ = \angle{B}$

$\sin^{-1} $\frac{BC}{AB}$ = \angle{A}$

$\angle{A} + \angle{B} + \angle{C} = \pi$

$\angle{C} = \frac{\pi}{2}$

$A + B = \frac{\pi}{2}$

As for the proof by identities, I haven't found it yet.

jcaron2 · Mar 4, 2011, 09:11 PM

Nice proof, Unknown008!