Trig - can't see how to simplify

[Brewer]

Homework Statement

Its all part of a show that question from an example paper that I'm doing. Apart from this part I have the answer. Just wondered if anyone could show me the substitution that I would have to make.

Starting from (\frac{sin3\gamma}{sin\gamma})^2, and ending up with (1+2cos2\gamma)^2

Other than this, I have the question right, and if I got to this stage in an exam, and couldn't do it after spending some time on it, then I'd just go to the final answer, and hope that its a logical step that I just haven't seen. But as I say I have the rest of the answer, its just this little part that needs manipulating.

Thanks

Brewer

 

[malawi_glenn]

hints:

sin(3y) = sin (2y + y) = .. ?

sin(2y) = .. ?

(cosy)^2 = .. ?

:)

 

[Hootenanny]

Since you've done the rest of the question, I see no harm is showing you this little bit. So, from the triple angle indentity \sin3\theta = 3\sin\theta - 4\sin^3\theta we can write;

\left(\frac{\sin3\gamma}{\sin\gammer}\right)^2 = \left(\frac{3\sin\gamma - 4\sin^3\gamma}{\sin\gamma}\right)^2

= \left(3-4\sin^2\gamma\right)^2

Recalling that \sin^2\theta = \left(1-\cos2\theta\right)/2 gives;

\left(3-4\sin^2\gamma\right)^2 = \left(3 - 4\frac{1-\cos2\gamma}{2}\right)^2

=\left(3 -\left(2 - 2\cos2\gamma\right)\right)^2

=\left(1+2\cos2\gamma\right)^2

As required. I hope that was helpful

 

[malawi_glenn]

my "solution" is simpler i guess..

i don't think this guy is at the level were one uses the triple angle indentity..

 

[Brewer]

It was actually. Both versions. I would have tried the first method, but a maths student I was talking to mentioned the triple angle identity (but I'm sure I've not heard of it/done it yet), but I didn't follow it. But I see how its done now. Thanks a lot, both of you.

 

[Hootenanny]

my "solution" is simpler i guess..

i don't think this guy is at the level were one uses the triple angle indentity..

An undergraduate doing a mathematics related course doesn't know the triple angle identity?

Of course the same result can be arrived using the angle sum and double angle indentities, but this way is quicker...

 

[cristo]

i don't think this guy is at the level were one uses the triple angle indentity..

The triple angle identity (which I admit, I don't remember explicitly) is easily derived by expanding sin(2x+x) and then using the double angle identities.

An undergraduate doing a mathematics related course doesn't know the triple angle identity?

Guilty as charged!

 

[Brewer]

An undergraduate doing a mathematics related course doesn't know the triple angle identity?

Of course the same result can be arrived using the angle sum indentities, but this was is quicker...

As far as I can remember, I haven't been taught it yet and I'm in my second year of a Physics degree. In fact I'm pretty definate that I haven't been taught it. I wasn't taught it last year, and my maths course this year focused mainly on Fourier transforms and series.

 

[malawi_glenn]

An undergraduate doing a mathematics related course doesn't know the triple angle identity?

Of course the same result can be arrived using the angle sum and double angle indentities, but this way is quicker...

depends on how much you must/want remember.. i think its easier to remember a few simple ones that have broad area of usage, than very very specific ones.

 

[Office_Shredder]

An undergraduate doing a mathematics related course doesn't know the triple angle identity?

If you've memorized triple angle identities, you're a loser

All statements made here reflect the opinions of the author and in no way reflect the views of normal people

 

[malawi_glenn]

If you've memorized triple angle identities, you're a loser

All statements made here reflect the opinions of the author and in no way reflect the views of normal people

naw that wasn't very nice =(

 

[Hootenanny]

Guilty as charged!

As far as I can remember, I haven't been taught it yet and I'm in my second year of a Physics degree. In fact I'm pretty definate that I haven't been taught it. I wasn't taught it last year, and my maths course this year focused mainly on Fourier transforms and series.

depends on how much you must/want remember.. i think its easier to remember a few simple ones that have broad area of usage, than very very specific ones.

If you've memorized triple angle identities, you're a loser

Guys, chill out a little! Obviously sarcasm doesn't work well over forums...

I wouldn't have known the triple angle forumlae either if I hadn't been using them a lot recently.

Office: Gee Thanks

 

[cristo]

Guys, chill out a little! Obviously sarcasm doesn't work well over forums...

Ahh.. so it was sarcastic! No, it doesn't work too well. There should be some kind of sarcastic smily on here! Anyway, I for one am totally chilled out!

You're right about the memorising thing though-- if you use something enough, you're bound to remember it!

 

[Office_Shredder]

Guys, chill out a little! Obviously sarcasm doesn't work well over forums...

I wouldn't have known the triple angle forumlae either if I hadn't been using them a lot recently.

Office: Gee Thanks

I was being sarcastic too (hence the italics)...

let's just pretend this conversation never happened

 

[Integral]

The triple angle identity is readily available in any decent table of trig identies. Once you get out of trig class, that is a perfectly valid reference which every student of math or physics should have at hand. You do not have to be "shown" each and every identity before you can use it. Learn to use references. My CRC Standard Math Tables is perhaps my most used reference.

 

[Curious3141]

The neatest way, IMHO, is to use factor formulae.

Forget about the squares on the LHS and RHS, and let's call gamma x since it saves typing.

\frac{\sin 3x}{\sin x} = \frac{2\sin 3x \cos x}{2 \sin x \cos x} = \frac{\sin 4x + \sin 2x}{\sin 2x} = \frac{\sin 2x(2\cos 2x + 1)}{\sin 2x} = 2\cos 2x + 1

BTW, Hootenanny, there's a little error in your working, and the final sign is wrong.

 

[Hootenanny]

BTW, Hootenanny, there's a little error in your working, and the final sign is wrong.

Duely corrected in the post, thanks Curious

This line;

=\left(3 -2 - 2\cos2\gamma\right)^2

Should read;

=\left(3 -\left(2 - 2\cos2\gamma\right)\right)^2

 

[Curious3141]

Duely corrected in the post, thanks Curious

This line;

=\left(3 -2 - 2\cos2\gamma\right)^2

Should read;

=\left(3 -\left(2 - 2\cos2\gamma\right)\right)^2

Much better!