How to get √(1-cosθ) from √(R² + R² − 2R² Cosθ)?

[Indranil]

How to solve it?
How to get √(1-cosθ) from √(R² + R² − 2R² Cosθ)?

 

[Harry_the_cat]

Can you give the whole question please?

 

[topsquark]

How to solve it?
How to get √(1-cosθ) from √(R² + R² − 2R² Cosθ)?

Well, your second expression is \(\displaystyle \sqrt{ R^2 + R^2 - 2R^2~cos( \theta )} = \sqrt{ R^2(1 + 1 - 2 ~ cos( \theta ))} = \sqrt{R^2} ~ \sqrt{2 - 2~cos( \theta )} = R \sqrt{2} ~ \sqrt{1 - cos( \theta )}\)
which is clearly not the same as \(\displaystyle \sqrt{1 - cos( \theta )}\)

So, as Harry the cat says, can you give us the whole question?

-Dan

 

[Indranil]

Well, your second expression is \(\displaystyle \sqrt{ R^2 + R^2 - 2R^2~cos( \theta )} = \sqrt{ R^2(1 + 1 - 2 ~ cos( \theta ))} = \sqrt{R^2} ~ \sqrt{2 - 2~cos( \theta )} = R \sqrt{2} ~ \sqrt{1 - cos( \theta )}\)
which is clearly not the same as \(\displaystyle \sqrt{1 - cos( \theta )}\)

So, as Harry the cat says, can you give us the whole question?

-Dan

I just wanted √(1-cosθ) solving the equation √(R² + R² − 2R² Cosθ). So I got it now.

 

[JeffM]

Well, your second expression is \(\displaystyle \sqrt{ R^2 + R^2 - 2R^2~cos( \theta )} = \sqrt{ R^2(1 + 1 - 2 ~ cos( \theta ))} = \sqrt{R^2} ~ \sqrt{2 - 2~cos( \theta )} = R \sqrt{2} ~ \sqrt{1 - cos( \theta )}\)
which is clearly not the same as \(\displaystyle \sqrt{1 - cos( \theta )}\)

So, as Harry the cat says, can you give us the whole question?

-Dan

Well, it is equal if [MATH]R = \dfrac{1}{\sqrt{2}}.[/MATH]

 

[topsquark]

I just wanted √(1-cosθ) solving the equation √(R² + R² − 2R² Cosθ). So I got it now.

What? I don't understand what you just said. And for the record you are not solving an equation \(\displaystyle \sqrt{R^2 + R^2 - 2R^2~cos( \theta )}\) is not an equation! Equations have equal signs in them.

You said you got the answer but just for the sake of curiousity what was the actual question?

-Dan