Functions - the fallacy in this argument

[autodidude]

I'm asked to find the fallacy in this argument

sin ≣ sin(π - x)
Hence, f(sin x) = f(sin(π-x))
Let f(x) = x sin x
Then x sin x = (π - x)sin(π-x)
x = π - x
Hence π = 2x and since x is any value we choose, so is π

The fourth line, where the author says the x in front of the sin x is also equal to π - x is where the error is, but in the answers, he says, ''the fallacy is we cannot choose f(sin x) to be x sin x, this is not a function of sin x alone but of x and sin x'

I suspect he's talking about the same thing, but I don't understand what it has to do with the fourth line.

 

[Disinterred]

From what I understand,

if you let

f(x) = xsinx

and u = sinx

then

f(u) = u*sin(u)

= sin(x)*sin(sin(x))

and likewise for sin(pi - x)

If you look at the proof, you see that the fourth line should be:

sin(x)*sin(sin(x)) = sin(pi - x)*sin(sin(pi - x))

which seems correct.

 

[autodidude]

So it's basically two functions in one?

 

[Mark44]

So it's basically two functions in one?

That's pretty meaningless in this problem.

The problem is that the 4th line does not evaluate f correctly. f maps a number into that number time the sin of that number.

For example,
f(x) = xsin(x)
f(y) = y sin(y)
f(a + b) = (a + b) sin(a + b)
f(sin(π-x)) = sin(π-x) sin(sin(π-x))

 

[JHamm]

Are you sure you typed it out correctly? Because the answers only make sense if either the function was defined as $f(\sin x) = x \sin x$ or if the author is deciding to be bizarrely pedantic about the scope of the variable x, otherwise the problem is definitely in the 4th line.

 

[nonequilibrium]

JHamm is correct, the author either made a typo or the OP did.

 

[autodidude]

That's pretty meaningless in this problem.

The problem is that the 4th line does not evaluate f correctly. f maps a number into that number time the sin of that number.

For example,
f(x) = xsin(x)
f(y) = y sin(y)
f(a + b) = (a + b) sin(a + b)
f(sin(π-x)) = sin(π-x) sin(sin(π-x))

Yeah, if I understand you correctly, that's what I thought. It's just his answer confused me.

@JHamm: Yeah, that's how the questioning and answer were given. I was wondering why he said we can't use f(sin x) to be x sin x because in the question he let f(x) = x sin x.

When I see f(sin x) = x sin x, I interpret that as sining the x before inputing it's sin(x)sin(sin(x)).

 

[JHamm]

If you take [itex]f(x) = x\sin x [/tex] you can then go
$$f(\sin x) = \sin x \sin \sin x$$
However if you try to define your function as
$$f(\sin x) = x\sin x$$
Then you're in trouble because that isn't surjective.