How Can Trig Double Angle Formulas Simplify This Function?

[TogoPogo]

Homework Statement

Hello, I am trying to simplify the inputted function here http://www.wolframalpha.com/input/?i=sqrt(2)+sqrt(1-cos((2pi(x-y))/n))

which is $\sqrt{2}$$\sqrt{1-cos[2\pi(x-y)/n]}$

to the form of 2sin[(x-y)\pi/n]

Homework Equations

Not sure

The Attempt at a Solution

EDIT: nevermind, found the solution... Forgot about double angle formula lolThe part that confused me was how they managed to remove the "2" in the argument of the cosine function, as well as how they removed the square root

 

[Redbelly98]

Let $z \equiv \pi (x-y) / n$

From the double-angle identity for cosine, we know that $\cos{(2z)} = 1 - 2 \sin^{2}{z}$, so
$$ \sqrt{2} \sqrt{1- \cos{(2z)}} = \sqrt{2} \sqrt{1 - (1 - 2 \sin^{2}{z})} $$
$$ = \sqrt{2} \sqrt{2 \sin^{2}{z}} = 2 \sqrt{\sin^{2}{z}} = 2 | \sin{z} | $$
Note the need for taking the absolute value in the final expression. (The initial expression always has a positive or zero value.)

 

[MidgetDwarf]

The last poster hit it on the nail.

I recommend to view and understand the proof of trigonometric identities that can be derived from 2 diagrams. The proof is really neat, as it involves using the distance formula 2 times.

Also, when you get an expression that has a lot going on, maybe try to substitute some values with dummy variables.

 

[member 587159]

The last poster hit it on the nail.

I recommend to view and understand the proof of trigonometric identities that can be derived from 2 diagrams. The proof is really neat, as it involves using the distance formula 2 times.

Also, when you get an expression that has a lot going on, maybe try to substitute some values with dummy variables.

What do you mean by 2 diagrams?

 

[haruspex]

Let $z \equiv \pi (x-y) / n$

From the double-angle identity for cosine, we know that $\cos{(2z)} = 1 - 2 \sin^{2}{z}$, so
$$ \sqrt{2} \sqrt{1- \cos{(2z)}} = \sqrt{2} \sqrt{1 - (1 - 2 \sin^{2}{z})} $$
$$ = \sqrt{2} \sqrt{2 \sin^{2}{z}} = 2 \sqrt{\sin^{2}{z}} = 2 | \sin{z} | $$
Note the need for taking the absolute value in the final expression. (The initial expression always has a positive or zero value.)

TogoPogo seems to have worked all that out already, nearly five years ago.

 

[MidgetDwarf]

What do you mean by 2 diagrams?

I will take a picture of the diagrams. Should have it posted by Thursday. It just involves 2 units circles.

 

[Redbelly98]

TogoPogo seems to have worked all that out already, nearly five years ago.

Indeed. But, with no replies and no posted solution, this thread had been placed in the "Open Problems" forum. Greg is encouraging people to post solutions to such threads:

Unanswered homework threads provide no value and can actually harm. Nearly 50% of traffic to PF lands on homework threads. If a guest lands on a thread with no replies it looks poorly and the guest is not helped. PF becomes a knowledge base when a thread is older than a few months.

Also, the answer stated in the OP was not completely correct, in missing the absolute value signs.

 

[haruspex]

Indeed. But, with no replies and no posted solution, this thread had been placed in the "Open Problems" forum. Greg is encouraging people to post solutions to such threads:

Also, the answer stated in the OP was not completely correct, in missing the absolute value signs.

Ok.