# Math Help - inverse sin etc.

1. ## inverse sin etc.

Ok i solved these problems.

a) $sin(arcsin(-\frac{1}{2}))$

$-\frac{1}{2}$ belongs to interval $[-\frac{\pi}{2]}, \frac{\pi}{2]}]$

so solution is $-\frac{1}{2}$

b) $arcsin(sin(\frac{7\pi}{6})$

$\frac{7\pi}{6}$ belongs to interval $\Re$
so solution is $\frac{7\pi}{6}$

c) $tg(arctg(-1))$

-1 belongs to $/Re$

so solution is -1

2. ## Re: Confirm if solutions are correct please.

a) and c) are correct but you can just use the fact that $\sin(\arcsin(x))=x$ and $\tan[\arctan(x)]=x$ (if you know why)
For b) first of all $\sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=\frac{-1}{2}$
So that means $\arcsin\left(\frac{-1}{2}\right) \in $\frac{-\pi}{2},\frac{\pi}{2}$$

So that means $\frac{7\pi}{6}$ can't never be an answer because it's not in $$\frac{-\pi}{2},\frac{\pi}{2}$$

3. ## Re: Confirm if solutions are correct please.

damn im confusing the domains

so for sin(arcsin) = x and arcsin(sin x) = x

x need to be in [-pi/2 , pi/2 ]

4. ## Re: Confirm if solutions are correct please.

You can't say $\arcsin[\sin(x)]=x$, question b is an example of why it's not true (also think about it why it's not true in general).
Actually I have given the answer to question b) with this quote:
Originally Posted by Siron
$\sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=\frac{-1}{2}$
That means $\arcsin\left(\frac{-1}{2}\right)=...$?

5. ## Re: Confirm if solutions are correct please.

Originally Posted by Siron
You can't say $\arcsin[\sin(x)]=x$, question b is an example of why it's not true (also think about it why it's not true in general).
Actually I have given the answer to question b) with this quote:

That means $\arcsin\left(\frac{-1}{2}\right)=...$?

$-\frac{\pi}{6}$

6. ## Re: Confirm if solutions are correct please.

Just more one thing.

sin(arcsin x ) = x if x belongs to interval [-1,1] right?

because i wrote in interval [-pi/2, pi/2].

7. ## Re: Confirm if solutions are correct please.

Originally Posted by Fabio010
Just more one thing.

sin(arcsin x ) = x if x belongs to interval [-1,1] right?

because i wrote in interval [-pi/2, pi/2].
Yes that's correct! Sorry, I hadn't seen that. Because the range of $y=\sin(x)$ is the closed interval [-1,1] that means the domain of (the inverse function) $y=\arcsin(x)$ is indeed the interval [-1,1].

And indeed $\frac{-\pi}{6}$ is correct.

8. ## Re: Confirm if solutions are correct please.

Ok thanks for the help !

9. ## Re: Confirm if solutions are correct please.

You're welcome!