Challenge your maths skills...
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.
Challenge your maths skills...
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.
You need to use two formulas...
$\displaystyle \sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}$
and
$\displaystyle \sin^2{x} + \cos^2{x} = 1$.
If you do some rearranging of the second, and some expanding of the first, it should fall into place...
Have you ever seen summation notation before?
Basically it's a sum, and the limits on the sum are your counter.
So here, $\displaystyle n = 13$.
So k will go between 0 and $\displaystyle \frac{13 - 1}{2} = 6$.
The $\displaystyle \left(_k^{n - k -1}\right)$ 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
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 $\displaystyle sin(3\theta)$
First step
$\displaystyle
\sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}$
Put n = 3 and accordingly k =1 so it becomes
$\displaystyle
\sin{(3x)} = \sin{x}\sum_{k = 0}^{\frac{3-1}{2}} (-1)^k \left(_k^{3 - k - 1}\right)2^{3 - 2k - 1}\cos^{3 - 2k - 1}{x}$
$\displaystyle
\sin{(3x)} = \sin{x}\sum_{k = 0}^{1} (-1)^k \left(_k^{3 - k - 1}\right)2^{3 - 2k - 1}\cos^{3 - 2k - 1}{x}$
Step 2
Expand the terms
$\displaystyle
\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}\}+$$\displaystyle \{ (-1)^1 \left(_1^{3 -1 - 1}\right)2^{(3 - (2 \times 1) - 1)}\cos^{(3 - (2\times 1) - 1)}{x}\}] $
--->$\displaystyle \left(_0^{3 - 0 - 1}\right) = \frac{2!}{(2-0)! \times 0!}$
$\displaystyle j! ~(spelled ~as~ j~ factorial)~ = j \times (j-1)\times (j-2)\times......1 $
Remember that
$\displaystyle 1!= 1 = 0!$
So
$\displaystyle \frac{2!}{(2-0)! \times 0!} = 1 $
Similarly
--->$\displaystyle \left(_1^{3 -1 - 1}\right) = \frac{1!}{0! \times 1!}=1 $
Step 3
Now gather the values of each term
First term
$\displaystyle
\{(-1)^0 \left(_0^{3 - 0 - 1}\right)2^{(3 - (2 \times 0) - 1)}\cos^{3 - 2\times 0 - 1}{x}\} = 1 \times 1 \times 4 \times cos^2 (x) = $$\displaystyle 4(1-sin^2(x))= 4 - 4sin^2(x) $
Second term
$\displaystyle \{ (-1)^1 \left(_1^{3 -1 - 1}\right)2^{(3 - (2 \times 1) - 1)}\cos^{(3 - (2\times 1) - 1)}{x}\} = -1\times 1 \times 1 \times 1 = -1$
Step 4
Add First and second term
$\displaystyle 3-4sin^2(x) $
Put this in the Step 2 which had those two terms
$\displaystyle
\sin{(3x)} = \sin{x}[3-4sin^2(x)] = 3sin(x)-4sin^3(x) $
Thus we have expressed sin(3x) in terms of $\displaystyle 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