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Proving tan(3x)

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Reply 20

Yes, but try not to change the form. Conclusion: my method was ackward and not very beneficial in a time-constrained situation.

Reply 21

tinypony

I did divide, but after that you end up with something nasty. Try it.

Reply 22

gyrase

chrisjorg

I did divide, but after that you end up with something nasty. Try it.

your method is fine, just when you divide you end up with some tans and some secs, you need to convert the secs to tans by using sec^2(x) = 1 + tan^2(x).

Reply 23

nota bene

chrisjorg

Yes, but try not to change the form.

I have not changed any form by multiplying by 1 as far as I am aware... :confused:

chrisjorg

Conclusion: my method was ackward and not very beneficial in a time-constrained situation.

Yes, your method is fine and works...

edit: seeing your next post... I have divided in my working up there and all vanishes nicely...

Reply 24

tinypony

your method is fine, just when you divide you end up with some tans and some secs, you need to convert the secs to tans by using sec^2(x) = 1 + tan^2(x).

Yes, but I did that. The problem was solved in the beginning of the thread.

--------------------------------------------------------------------------------

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.
:-)

Reply 25

gyrase

chrisjorg

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'll use these identities which are easily derived from De's theorem.

then,
tan(3x) = sin(3x)/cos(3x)
= (3sinx-4sin^3x)/(4cos^3x-3cosx)
= (3tanxsec^2x-4tan^3x)/(4-3sec^2x)
= (3tanx(1+tan^2x)-4tan^3x)/(4-3(1+tan^2x))
= (3tanx-tan^3x)/(1-3tan^2x)

Now accept that, I'm not up for mind-games.

Reply 26

nota bene

I'll just show what I mean as you still seem to think you get something messy :s-smilie:

(or am I missing something?)

tan(3a)=\frac{sin(3a)}{cos(3a)}=\frac{3cos^2(a)sin(a)-sin^3(a)}{cos^3(a)-3cos(a)sin^2(a)}

and multiplying by

\frac{\frac{1}{cos^3(a)}}{\frac{1}{cos^3(a)}}

Gives

\frac{\frac{3cos^2(a)sin(a)}{cos^3(a)}-\frac{sin^3(a)}{cos^3(a)}}{\frac{cos^3(a)}{cos^3(a)}-\frac{3cos(a)sin^2(a)}{cos^3(a)}}

. This gives (cancelling cos everywhere and then converting the remaining to tan(a)) :

\frac{3tan(a)-tan^3(a)}{1-3tan^2(a)}

Reply 27

tinypony

NB, you need to use the method Decota outlined (which I was discussing in the beginning of the thread).

then,
tan(3x) = sin(3x)/cos(3x)
= (3sinx-4sin^3x)/(4cos^3x-3cosx)
= (3tanxsec^2x-4tan^3x)/(4-3sec^2x) ***
= (3tanx(1+tan^2x)-4tan^3x)/(4-3(1+tan^2x))
= (3tanx-tan^3x)/(1-3tan^2x)

The messy part is ***, because you need to change it to sec^2x and then 1+tan^2x...

Reply 28

tinypony

NB your method is obviously more simple, but as I have been stressing it is imperative that you use the form (3sinx-4sin^3x)/(4cos^3x-3cosx) because the question says "hence"

Reply 29

DFranklin

nota bene

I'll just show what I mean as you still seem to think you get something messy :s-smilie:

(or am I missing something?)
As far as I can see, you are 100% correct.

As far as arguing about "hence", since the required identity comes out in a single line (with the explanation "divide top and bottom by cos^3 x), my feeling is you don't get much more "hence" than that!

Reply 30

nota bene

chrisjorg

NB your method is obviously more simple, but as I have been stressing it is imperative that you use the form (3sinx-4sin^3x)/(4cos^3x-3cosx) because the question says "hence"

Okay, well it is me interpreting that 'hence' a bit differently...
I reasoned like "if a=b=c and d=e=f and hence prove c/f=yadda then that implies I can use b/e=yadda", which I don't know how valid it is.
Anyway, we both got there, and I know how you did it, I've just been trying to figure out what was wrong with my way :wink:

(i.e. the 'hence' which I think I am using in an okay way, though I may be wrong there...)

edit: just saw your reply David, and I'm a bit relieved you agree with me... :smile:

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