1. ## trigonometric expansion

Question:
Express sin13θ in the form
Asinθ + Bsin^3θ + Csin^5θ +Dsin^7θ + Esin^9θ +Fsin11^θ + Gsin^13θ
A to G being natural numbers.

2. Originally Posted by cogito_ergo_sum

Question:
Express sin13θ in the form
Asinθ + Bsin^3θ + Csin^5θ +Dsin^7θ + Esin^9θ +Fsin11^θ + Gsin^13θ
A to G being natural numbers..

Its up to you to show us what you got!!
Help... this is too hard for me...

i think i need to use complex numbers somewhere... help me, maths buddies!
You need to use two formulas...

$\sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}$

and

$\sin^2{x} + \cos^2{x} = 1$.

If you do some rearranging of the second, and some expanding of the first, it should fall into place...

3. Originally Posted by Prove It
You need to use two formulas...

$\sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}$

and

$\sin^2{x} + \cos^2{x} = 1$.

If you do some rearranging of the second, and some expanding of the first, it should fall into place...
Can you please explain how to use the first formula... i am a bit confused. tahnks a lot though

4. Originally Posted by cogito_ergo_sum
Can you please explain how to use the first formula... i am a bit confused. tahnks a lot though
Have you ever seen summation notation before?

Basically it's a sum, and the limits on the sum are your counter.

So here, $n = 13$.

So k will go between 0 and $\frac{13 - 1}{2} = 6$.

The $\left(_k^{n - k -1}\right)$ is a term in Pascal's triangle.

It might help if you did some research on summation (sigma) notation and Pascal's Triangle.

Summation - Wikipedia, the free encyclopedia

Pascal's triangle - Wikipedia, the free encyclopedia

5. Originally Posted by Prove It
Have you ever seen summation notation before?

Basically it's a sum, and the limits on the sum are your counter.

So here, $n = 13$.

So k will go between 0 and $\frac{13 - 1}{2} = 6$.

The $\left(_k^{n - k -1}\right)$ is a term in Pascal's triangle.

It might help if you did some research on summation (sigma) notation and Pascal's Triangle.

Summation - Wikipedia, the free encyclopedia

Pascal's triangle - Wikipedia, the free encyclopedia

I did exactly that and it doesn't work out properly. I used the binomial theorem and still stuck.
Thanks a lot for the help but i still don't know where i went wrong?

6. Originally Posted by cogito_ergo_sum
I did exactly that and it doesn't work out properly. I used the binomial theorem and still stuck.
Thanks a lot for the help but i still don't know where i went wrong?
First of all you should thank Prove It that he showed you the way rather than giving you the answer directly and that's the real spirit of MHF

Now it comes to your question , since you are still unable to catch with it , I will give you an example along with the step numbers**

example for $sin(3\theta)$

First step

$
\sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}$

Put n = 3 and accordingly k =1 so it becomes

$
\sin{(3x)} = \sin{x}\sum_{k = 0}^{\frac{3-1}{2}} (-1)^k \left(_k^{3 - k - 1}\right)2^{3 - 2k - 1}\cos^{3 - 2k - 1}{x}$

$
\sin{(3x)} = \sin{x}\sum_{k = 0}^{1} (-1)^k \left(_k^{3 - k - 1}\right)2^{3 - 2k - 1}\cos^{3 - 2k - 1}{x}$

Step 2

Expand the terms

$
\sin{(3x)} = \sin{x}[ \{(-1)^0 \left(_0^{3 - 0 - 1}\right)2^{(3 - (2 \times 0) - 1)}\cos^{(3 - (2\times 0) - 1)}{x}\}+$
$\{ (-1)^1 \left(_1^{3 -1 - 1}\right)2^{(3 - (2 \times 1) - 1)}\cos^{(3 - (2\times 1) - 1)}{x}\}]$

---> $\left(_0^{3 - 0 - 1}\right) = \frac{2!}{(2-0)! \times 0!}$

$j! ~(spelled ~as~ j~ factorial)~ = j \times (j-1)\times (j-2)\times......1$

Remember that

$1!= 1 = 0!$

So

$\frac{2!}{(2-0)! \times 0!} = 1$

Similarly
--->
$\left(_1^{3 -1 - 1}\right) = \frac{1!}{0! \times 1!}=1$

Step 3

Now gather the values of each term

First term

$
\{(-1)^0 \left(_0^{3 - 0 - 1}\right)2^{(3 - (2 \times 0) - 1)}\cos^{3 - 2\times 0 - 1}{x}\} = 1 \times 1 \times 4 \times cos^2 (x) =$
$4(1-sin^2(x))= 4 - 4sin^2(x)$

Second term
$\{ (-1)^1 \left(_1^{3 -1 - 1}\right)2^{(3 - (2 \times 1) - 1)}\cos^{(3 - (2\times 1) - 1)}{x}\} = -1\times 1 \times 1 \times 1 = -1$

Step 4

$3-4sin^2(x)$

Put this in the Step 2 which had those two terms

$
\sin{(3x)} = \sin{x}[3-4sin^2(x)] = 3sin(x)-4sin^3(x)$

Thus we have expressed sin(3x) in terms of
$Asin(x) + Bsin^3(x)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
**Tell the step number you didn't get

-All you need to do is do the same way for sin(13x) find k, expand,find term values and go ahead