# tan(3x) in terms of tan(x)

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#1
I've already done most of the working out for this, which I won't bother posting as my answer agrees with the back of the book. But this is where I'm suck:

I know that
and

The next step would then be to say that

...but now what? You could take tan(x) out of the fraction, but I still don't know how to go about simplifying it.

The book says the answer is

+rep is available
1
12 years ago
#2
Rewrite sin^3 in terms of sin and cos^2

Then divide top and bottom by cos^3.
0
12 years ago
#3
can't you use the addition or double angle formula for tan?
0
#4
(Original post by DFranklin)
Rewrite sin^3 in terms of sin and cos^2

Then divide top and bottom by cos^3.
Ah yes, I've got it now, thanks.

And I probably could have used the compound angle formula somehow, except for the slight problem that I haven't done C4 yet . Doing FP1 before C1, then FP2 before C3 is fun
0
12 years ago
#5
If you're doing this by de Moivre, the trick is to keep the form you get from initially expanding (C+iS)^3, (where C = cos x, S = sin x) rather than rewriting to get sin 3x in terms of only sin x.

i.e.

(C+iS)^3 = C^3+ 3i C^2S - 3 C S^2 - iS^3.

So sin 3x =3 C^2 S - S^3, cos 3x = C^3-3CS^2.
And .

Then just divide by C^3 to rewrite in terms of tan x.
0
12 years ago
#6
(Original post by tomthecool)
Ah yes, I've got it now, thanks.

And I probably could have used the compound angle formula somehow, except for the slight problem that I haven't done C4 yet . Doing FP1 before C1, then FP2 before C3 is fun
How come you did that? We've only just finished FP1 having done C1-4 last year.
0
#7
(Original post by joshm)
How come you did that? We've only just finished FP1 having done C1-4 last year.
I'm doing maths and further maths at the same time... which is mostly OK, but occasionally screws things up.

For example, I had to pretty much learn all the basic calculus stuff when doing M1 before C2. (And didn't do any calculus in C1.) This year, I'm meant to be doing some much harder calculus stuff in FP2, but I started the chapter before even touching C3 calculus.
And it gets even worse with other modules... I did S2 before S1 (on 1 lesson per week!), which meant I had to really rush through the basics of binomial distributions and stuff in order to understand the more complex S2 distributions. I did FP1 with less than 2 lessons per week (and no textbook lol), with loads of previous knowledge missing from other core mosules...yet still somehow got 100% on the exam!

But looking on the bright side, all the disadvantages I've been through last year, combined with pretty good results, will hopefully make the unis I applied to love me
0
12 years ago
#8
(Original post by tomthecool)
I'm doing maths and further maths at the same time... which is mostly OK, but occasionally screws things up.

For example, I had to pretty much learn all the basic calculus stuff when doing M1 before C2. (And didn't do any calculus in C1.) This year, I'm meant to be doing some much harder calculus stuff in FP2, but I started the chapter before even touching C3 calculus.
And it gets even worse with other modules... I did S2 before S1 (on 1 lesson per week!), which meant I had to really rush through the basics of binomial distributions and stuff in order to understand the more complex S2 distributions. I did FP1 with less than 2 lessons per week (and no textbook lol), with loads of previous knowledge missing from other core mosules...yet still somehow got 100% on the exam!

But looking on the bright side, all the disadvantages I've been through last year, combined with pretty good results, will hopefully make the unis I applied to love me
Jesus christ. Good luck.
0
8 years ago
#9
tan(A +B) = tanA + tanB/(1- tanA*tanB) .........................(1)

well u can use dis formula
let A=2x B=x

Put A = B = x in (1)
We get tan(2x) = 2*tan(x)/(1-tan^2 (x)) ...........................(2)

Now put A = 2x, B=x in (1) ,using relation (2).
We get tan(3x) = [2tan(x)/(1-tan^2(x)) + tan(x)]/[ 1 - {2tan^2(x)}/{1- tan^2(x)}]
Simplifying( I know this looks tough but put pen to paper - and you'll see how simple it is.),
we get

tan(3x) = [3 tan(x) - tan^3(x)]/[1 - 3 * tan^2(x)]
0
3 years ago
#10
i think he denominator is wrongThe book says the answer is \tan(3x)=\frac{3\tan(x)-\tan^3(x)}{3-3\tan^2(x)}
0
3 years ago
#11
(Original post by ytj)
i think he denominator is wrongThe book says the answer is \tan(3x)=\frac{3\tan(x)-\tan^3(x)}{3-3\tan^2(x)}
0
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