1. ## Simplifying/Solving Trig Function

I got this problem on a test. The book didn't cover it (and I would not suggest this Trig text to anyone, the examples are horrible) and my teacher didn't cover it. These types of problems weren't even in the chapter test from the book so you can see how I was so distressed when I was tested on it. But I digress...

Simplify:

arcsin(sin x) if pi/2 =< x =< pi

2. Originally Posted by symstar
I got this problem on a test. The book didn't cover it (and I would not suggest this Trig text to anyone, the examples are horrible) and my teacher didn't cover it. These types of problems weren't even in the chapter test from the book so you can see how I was so distressed when I was tested on it. But I digress...

Simplify:

arcsin(sin x) if pi/2 =< x =< pi
if you sine something and then arcsine it, you get the original thing again.

arcsin(sin x) = x if pi/2 =< x < pi

arcsin(sin x) = 0 if x = pi

i have two answers above, why? we have to account for the fact that our calculators always takes us back to the reference angle, which is not always the same as the original angle.

so in general, for this problem, we should say something like:

$\displaystyle \arcsin \left( \sin x \right) = x \pm \frac {k \pi}{2}$ for some integer $\displaystyle 0 \leq k \leq 2$

and we choose the appropriate k

3. How do the restrictions come into play on problems similar to this?

4. Because

arcsin(sin x) (not =) x if x is outside that restriction.

Take for example x = 2*pi instead of 0 then.

arcsin (sin 2pi) = arcsin (0) = 0

See that 2pi and 0 are not the same.