# Injective function

Tags:
1. Mar 16, 2015

1. The problem statement, all variables and given/known data

Prove that sinx+cosx is not one-one in [0,π/2]
2. Relevant equations

None
3. The attempt at a solution

Let f(α)=f(β)
Then sinα+cosα=sinβ+cosβ
=> √2sin(α+π/4)=√2sin(β+π/4)
=> α=β
so it has to be one-one

2. Mar 16, 2015

### Dick

The sine function is not one-one. You can't just invert it without thinking about the domain. Try x=0 and x=pi/2. What do you say now?

3. Mar 16, 2015

### Raghav Gupta

You can also use calculus.
Differentiate the given function.
It must always have a positive or negative slope but not both.

4. Mar 17, 2015

I know that. But the result I got is contradicting.

5. Mar 17, 2015

### Raghav Gupta

As dick said, you can't take inverse simply like that. Your 3rd step is wrong and that's why your result is wrong.
The inverse taking depends on domain.
For example f(x) = x2 is not one-one for domain R.
It might seem so.
x12= x22
Taking square root on both sides.
X1= x2
But that's not correct.

6. Mar 17, 2015

Why cant I take inverse diretly?
In your example, I know that x1=x2 or -x2
but here you can say that from 0 to pi/4, x1=x2

7. Mar 17, 2015

### Raghav Gupta

Yes here 0 to pi/4 your function is one-one but 0 to pi/2 it is not one-one.
See this,
Log 0 is not defined and taking inverse or anti log of it would seem to someone it's zero. But we cannot take here simply the inverse.
So for particular domains we can't take the inverse.