Proving sinx+cosx is not one-one in [0,π/2]

[AdityaDev]

Homework Statement

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

Homework Equations

None

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
[/B]

 

[Dick]

Homework Statement

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

Homework Equations

None

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[/B]

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?

 

[Raghav Gupta]

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

 

[AdityaDev]

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?

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

 

[Raghav Gupta]

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

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.

 

[AdityaDev]

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.

Why can't 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

 

[Raghav Gupta]

Why can't 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

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.