Hello, I'm stuck with an trig identity

[saaz]

Hello, I'm stuck with an trig identity, can someone help me? Thanks in advance ?

[math]-cot(x/2) = (sin2x+sinx)/(cos2x-cosx)[/math]
Saaz

 

[jonah2.0]

Beer induced reaction follows.

Hello, I'm stuck with an trig identity, can someone help me? Thanks in advance ?

[math]-cot(x/2) = (sin2x+sinx)/(cos2x-cosx)[/math]
Saaz

It's not an identity; you'll be stuck forever.

saaz Hello, I'm stuck with an trig identity

www.desmos.com

 

[saaz]

Really????????? Are sure? Fu**ing books that tell to verify each identity and don't say in the solutions that it is not. I've spent 4 hours on this thing. Jeez.

How did you get so quickly that it isn't an identity? Wolframalpha says this:

 

[saaz]

I actually checked on a graphic calculator. It is an identity ?

 

[jonah2.0]

Beer induced reaction follows.

Really????????? Are sure? Fu**ing books that tell to verify each identity and don't say in the solutions that it is not. I've spent 4 hours on this thing. Jeez.

How did you get so quickly that it isn't an identity? Wolframalpha says this: View attachment 33520View attachment 33520

The Wolfram|Alpha god is clearly more hammered than I am.
As you can see with the Desmos demigod, the red and blue graphs don't match.

Edit: Btw, what book wrere you using?

 

[saaz]

I checked on desmos and the graphs match ?

 

[BigBeachBanana]

The Wolfram|Alpha god is clearly more hammered than I am.
As you can see with the Desmos demigod, the red and blue graphs don't match.

There's an extra x in your equation. The very last one on the bottom. The identity is in fact true.

 

[Deleted member 4993]

Hello, I'm stuck with an trig identity, can someone help me? Thanks in advance ?

[math]-cot(x/2) = (sin2x+sinx)/(cos2x-cosx)[/math]
Saaz

sin(2x) + sin(x) = sin(3x/2 + x/2) + sin(3x/2 - x/2) = 2 * sin(3x/2) * cos(x/2)......edited

cos(2x) - cos(x) → -2 * sin(3x/2) * sin(x/2)

[sin(2x) + sin(x)] / [cos(2x) - cos(x)] = [2 * sin(3x/2) * cos(x/2)] / [-2 * sin(3x/2) * sin(x/2)] = - cot(x/2).......edited

 

[jonah2.0]

Beer induced reaction follows.

There's an extra x in your equation. The very last one on the bottom. The identity is in fact true.

View attachment 33522

Aye.
My bad.
Drunk finger.

 

[Harry_the_cat]

I'd start off by using the double angle formulae to rewrite sin(2x) and cos(2x) in terms of sin x and cos x.
Factorise and cancel.
Then use the double angle formulae again to write sin(x) and cos(x) in terms of sin(x/2) and cos(x/2).
Cancel and it all falls out.
Show us what you have done.

 

[saaz]

Thank you all! I think I have to start again the chapter...

Meanwhile

sin(2x) + sin(x) = sin(3x/2 + x/2) + sin(3x/2 - x/2) = 2 * sin(3x/2) * sin(x/2)

cos(2x) - cos(x) → -2 * sin(3x/2) * sin(x/2)

[sin(2x) + sin(x)] / [cos(2x) - cos(x)] = [2 * sin(3x/2) * sin(x/2)] / [-2 * sin(3x/2) * sin(x/2)] = - cot(x/2)

Thanks! this is black magic. Can you tell me how did you get the last Sin(x/2) of the first line? Because from the double angle identity it comes out as a cos(x/2)...
Second, I don't see how this [2 * sin(3x/2) * sin(x/2)] / [-2 * sin(3x/2) * sin(x/2)] is equivalent to the - cot(x/2). I also checked in the graphic calculator, but it is not.

Anyway, now that I use the sum to product identity it's actually quite simple...

I'd start off by using the double angle formulae to rewrite sin(2x) and cos(2x) in terms of sin x and cos x.
Factorise and cancel.
Then use the double angle formulae again to write sin(x) and cos(x) in terms of sin(x/2) and cos(x/2).
Cancel and it all falls out.
Show us what you have done.

I was using the double angle formulae, I factorised the nominator, but I didn't manage to find a way to factorise the denominator (after using the double angle identity to cos(2x).

 

[BigBeachBanana]

Beer induced reaction follows.
Aye.
My bad.
Drunk finger.

Perhaps you were seeing double, a common side effect of alcohol.

 

[Harry_the_cat]

\(\displaystyle cos(2x) - cos(x) = 2 cos^2(x) -1 - cos(x) =2cos^2(x) - cos(x) -1\)
Letting \(\displaystyle a=cos(x)\), you have \(\displaystyle 2a^2-a-1\). Can you factorise that?

 

[saaz]

\(\displaystyle cos(2x) - cos(x) = 2 cos^2(x) -1 - cos(x) =2cos^2(x) - cos(x) -1\)
Letting \(\displaystyle a=cos(x)\), you have \(\displaystyle 2a^2-a-1\). Can you factorise that?

Ooh, now I see: (2cos(x)+1)(cos(x)-1).

Thanks again, you were super helpful.

 

[jonah2.0]

Beer soaked ramblings follow.

Perhaps you were seeing double, a common side effect of alcohol.

Definitely.
Me beer goggles can be very untrustworthy when I'm lying down on me bed with my phone clamped with one of them convenient phone holder thingamajigs. Good way to fall sleep while watching YouTube vids. Especially when I'm watching one them uploaded Ancient Aliens episodes; the narrator's voice really lulls me to sleep even without beer in me veins.

 

[Steven G]

After 4 hours (actually sooner) I would get suspicious that this was not an identity and plug in random values for x and see if both sides are equal. If they are not, then it is in fact not an identity and there is no need to continue. If on the other hand, you keep getting equality then you could assume that it is an identity and continue on.