Trigonomitry

[TSN79]

I'm a bit confused at the moment. All my books say that normal and inverse trig.functions cancel eachother out like this

\sin(\arcsin(x))=x and \arcsin(sin(x))=x

But when I try this out on my calculator - TI-89 - it only wants to recognize the first equation as being equal to x. Is that true or...? :confused:

[Galileo]

For a function to have an inverse, it needs to be injective and surjective.
The problem is that sin(x) is generally not injective, however it is injective on (eg) the interval [-\pi/2,\pi/2]. This interval can be chosen as the domain of sin(x) and becomes the range of arcsin(x), so for x in this interval the second equation will hold as well.

[TSN79]

But since my calculator always accepts sin(arcsin(x)) as being equal to x, does that mean that this equation is always true no matter what? :confused:

[Galileo]

That's funky. You cannot take the arcsin of x if |x|>1, since such an x wouldn't be in the range of sin(x).
My calculator gives an error, as it should.
(Unless you use complex numbers).

[krab]

Think it through. Try x=360 degrees. sin(360 deg.)=0, but arcsin(0)=0. So
arcsin(sin(x)) is not equal to x. OTOH, sin(arcsin(x))=x

[HallsofIvy]

Another function that does not have a "true" inverse is f(x)= x2 and you might find it easier to look at that function (you can do the calculations yourself rather than use a calculator).

If we restrict the domain to non-negative numbers, THEN we can use f-1(x)= √(x) as the inverse.

√(x) only makes sense when x is positive. For any positive number, if you first take the square root and then square, you get x back again: f(f-1(x))= x.
In particular, if x= 4, f-1(4)= √(4)= 2 and f(2)= 22= 4 again.

However, f(x) is defined for x positive or negative. If x= 2, f(2)= 22= 4 and then f-1(4)= √(4)= 2. In that case, f[sup]-1[\sup](f(x))= x.

But if x= -2, f(-2)= (-2)2= 4 and then f-1(4)= √(4)= 2.
In that case, f-1(f(x)) is NOT equal to x.

[tmwong]

sin(arcsin(x))=x true while arcsin(sin(x))=x is not true.

[Galileo]

sin(arcsin(x))=x does not make sense if |x|>1, since it is undefined.
It's only true if |x|<=1.

[TSN79]

So how would I go about it to find the solutions for let's say tan(x/2)=0 ?
I thought I could make it easier by using arctan(tan(x/2))=x/2, but you're telling me that this wouldn't work? What if I decided to look only at a part of tan(x/2) that is increasing or decreasing? Let's say from -pi to pi. Would this work? Because that's the requirement for a trig.function to have an inverse isn't it? Would arctan(tan(x/2))=x/2 be true then?

[Galileo]

tan(x) has the same problem: It isn't injective.
However, it IS injective if it is restricted to the interval (-\frac{\pi}{2},\frac{\pi}{2})
Then for x in this interval \arctan(\tan(x))=x holds.
(Notice that range of arctan(x) is (-\frac{\pi}{2},\frac{\pi}{2}))

If you wish to solve \tan(x/2)=0, then using arctan will give you an answer that lies in the interval (-\frac{\pi}{2},\frac{\pi}{2}). (In this case x=0).
For a complete answer, you need additional info.
In this case you can use that the tangent is periodic (period \pi)
so that x/2=k\pi where k is an integer will give the general solution).

[TSN79]

Wouldn't period of tan(x/2) be 2pi? When the interval of tan(x) is restricted to <-pi/2 , pi/2>, the range of arctan(x) is the same. But when the interval of tan(x/2) is <-pi , pi>, but now the range of arctan(x/2) is not the same. Why is this? Shouldn't the range of arctan(x/2) the same as the interval of tan(x/2)?

[Galileo]

Yes, the period of tan(x/2) is 2\pi (I meant the period of tan(x) in my post).

If y=\tan(x/2). Then for x \in (-\pi,\pi):
\arctan(y)=x/2
x=2\arctan(y)
So the inverse function of tan(x/2) is 2arctan(x) and has range (-\pi,\pi)