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

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

This discussion is closed.

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

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

(Original post by

Rewrite sin^3 in terms of sin and cos^2

Then divide top and bottom by cos^3.

**DFranklin**)Rewrite sin^3 in terms of sin and cos^2

Then divide top and bottom by cos^3.

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

Report

#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.

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

Report

#6

(Original post by

Ah yes, I've got it now, thanks.

And I probably

**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**
0

(Original post by

How come you did that? We've only just finished FP1 having done C1-4 last year.

**joshm**)How come you did that? We've only just finished FP1 having done C1-4 last year.

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

Report

#8

(Original post by

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

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

**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

0

Report

#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)]

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

Report

#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

Report

#11

(Original post by

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)}

**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

X

Page 1 of 1

Go to first unread

Skip to page:

new posts

Back

to top

to top