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Proving tan(3x)
Hi,
I was deducing the formula for tan(3x) using de Moivre's theorem. So first I deduced the formulas for sin(3x) and cos(3x) by equating real and imaginary parts. Then they ask me to deduce that tan(3x) = [3tanx-tan^3x]/[1-3tan^2x]
Can this be done by simply manipulating (sin3x)/(cos3x), because I do not see how this could be done. I do not want it proven using the fact that tan3x=tan(2x+x) as this has nothing to do with complex numbers.
Any ideas?
I was deducing the formula for tan(3x) using de Moivre's theorem. So first I deduced the formulas for sin(3x) and cos(3x) by equating real and imaginary parts. Then they ask me to deduce that tan(3x) = [3tanx-tan^3x]/[1-3tan^2x]
Can this be done by simply manipulating (sin3x)/(cos3x), because I do not see how this could be done. I do not want it proven using the fact that tan3x=tan(2x+x) as this has nothing to do with complex numbers.
Any ideas?
Scroll to see replies
http://www.thestudentroom.co.uk/showthread.php?t=379330&highlight=complex+analysis
Does this help? (I did the tan(4x) here so should hint you...)
edit: well, I imagine you should use divide the numerator and the denominator by cos^3(x) this time...
Does this help? (I did the tan(4x) here so should hint you...)
edit: well, I imagine you should use divide the numerator and the denominator by cos^3(x) this time...
Reply 2
tinypony
OP
Yes, I had done that before but it yields a mess!
I got it now, the trick is to make 1/cos^2x = sec^2x
Now sec^2x = 1+tan^2x and manipulating does indeed yield the identity.
:-)
I got it now, the trick is to make 1/cos^2x = sec^2x
Now sec^2x = 1+tan^2x and manipulating does indeed yield the identity.
:-)
chrisjorg
I got it now, the trick is to make 1/cos^2x = sec^2x
Now sec^2x = 1+tan^2x and manipulating does indeed yield the identity.
:-)
Now sec^2x = 1+tan^2x and manipulating does indeed yield the identity.
:-)
I think you are making it unnecessarily complicated by having to use that identity...
If I am not missing something you can do like this (which to me seems simpler than what you suggested):
Spoiler
this all took me about 2-3 minutes on paper, so shouldn't be that demanding... (I stumbled upon sin^9(x) last week, NOT nice to come up on an exam)
Reply 4
tinypony
OP
Yes but the question required deriving sin(3x) and cos(3x) by means of de Moivre:
cis3x=cos(3x)+isin(3x)=......=something real + i(something imaginary)
Then equating real and imaginary parts you get the desired identities, then they ask you to "hence" derive tan(3x).
It was simply the messed up forms of sin(3x) and cos(3x) that made it seem impossible.
Yes, it would be a lot easier to either use the binomial theorem or simply tan(2x+x)...
cis3x=cos(3x)+isin(3x)=......=something real + i(something imaginary)
Then equating real and imaginary parts you get the desired identities, then they ask you to "hence" derive tan(3x).
It was simply the messed up forms of sin(3x) and cos(3x) that made it seem impossible.
Yes, it would be a lot easier to either use the binomial theorem or simply tan(2x+x)...
chrisjorg
Yes but the question required deriving sin(3x) and cos(3x) by means of de Moivre:
cis3x=cos(3x)+isin(3x)=......=something real + i(something imaginary)
Then equating real and imaginary parts you get the desired identities, then they ask you to "hence" derive tan(3x).
It was simply the messed up forms of sin(3x) and cos(3x) that made it seem impossible.
cis3x=cos(3x)+isin(3x)=......=something real + i(something imaginary)
Then equating real and imaginary parts you get the desired identities, then they ask you to "hence" derive tan(3x).
It was simply the messed up forms of sin(3x) and cos(3x) that made it seem impossible.
I am using de moivre to get to that...
. Have you looked through my working? (there may be something wrong, but I can't spot it).I see how this can be done in more ways, but I am just thinking that this is the quickest way I see...
Reply 6
tinypony
OP
Yes but the identities used were:
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
I think it is because these are slight modifications.
Your identites are a lot more obvious as they yield the identies without further modifications. Try the ones I listed, when you do so you will be wondering how the **** you remove the coefficients
The IBID book can be annoying as their questions are tricky...
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
I think it is because these are slight modifications.
Your identites are a lot more obvious as they yield the identies without further modifications. Try the ones I listed, when you do so you will be wondering how the **** you remove the coefficients
The IBID book can be annoying as their questions are tricky...
Reply 9
tinypony
OP
Well, I derived them from de Moivre in my 1st and second line of working...
Yes and that is also a lot easier, but the questions goes as follows:
Deduce that cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
using de Moivre. Hence deduce that tan(3x)....
Reply 10
tinypony
OP
Thanks but it was derived using the same method. Although realizing the thing with secant did take some time...
Reply 11
tinypony
OP
The hypergeometric distribution, is that in the core?
chrisjorg
Yes and that is also a lot easier, but the questions goes as follows:
Deduce that cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
using de Moivre. Hence deduce that tan(3x)....
Deduce that cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
using de Moivre. Hence deduce that tan(3x)....
I interpret that as I am instructed to use the earlier results, that is the expansion for cos(3x) and sin(3x) and therefore my working would satisfy the question
?I'm just discussing this because it feels important to know I can interpret questions
.chrisjorg
Thanks but it was derived using the same method. Although realizing the thing with secant did take some time...
sorry check this http://thestudentroom.co.uk/showthread.php?t=362560&highlight=tan3x
Reply 15
tinypony
OP
What? since when :P
The H&H book has nothing on that type of distribution....
No, hence means use:
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
and nothing else.
The H&H book has nothing on that type of distribution....
No, hence means use:
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
and nothing else.
chrisjorg
What? since when :P
The H&H book has nothing on that type of distribution....
No, hence means use:
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
and nothing else.
The H&H book has nothing on that type of distribution....
No, hence means use:
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
and nothing else.
Hmm, H&H sucks? (jk, but seriously have a look in the syllabus then... it is on p. 539 in IBID core...)
Yes, of course hence means that! it is the same as I am saying, and then naturally since tan is defined as sin/cos we can write tan(3x)=sin(3x)/cos(3x) and work on as I have done
chrisjorg
What? since when :P
The H&H book has nothing on that type of distribution....
No, hence means use:
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
and nothing else.
The H&H book has nothing on that type of distribution....
No, hence means use:
cos3x=4cos^3x-3cosx
sin3x=3sinx-4sin^3x
and nothing else.
Are you saying you don't know that tan(3x) = sin(3x)/cos(3x) or that you don't know how to divide top and bottom by cos^3(x)?
Reply 18
tinypony
OP
Lol :-) major communication issues here.
I was saying you need to use the RHS of both identities: 4cos^3x-3cosx
3sinx-4sin^3x
Try dividing them, it gets messy.
I was saying you need to use the RHS of both identities: 4cos^3x-3cosx
3sinx-4sin^3x
Try dividing them, it gets messy.
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