How to find $K$ in $2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$? The problem is as follows:
Which value should have $K$ in the expression from below so it makes an identity?
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&11\\
2.&12\\
3.&14\\
4.&15\\
\end{array}$
I'm confused exactly how to get that particular value for $K$.
The only thing which I was able to spot is that: (Using manipulations on the left side of the equation)
$2\csc 30 ^\circ(\cos 8x+\cos 6x)(\cos 6x + \cos 2x)-1=\sin (Kx)\cdot \csc x$
$2\csc 30 ^\circ (2\cos 7x \cos x)(2\cos 4x \cos 2x)-1$
$4 (2\cos 7x \cdot \cos 2x)(2 \cos x \cdot 4x)-1$
$4(\cos 9x + \cos 5x)(\cos 5x+\cos 3x)-1$
$4(\cos 9x \cos 5x +\cos 9x \cos 3x + \cos 5x \cos 5x + \cos 5x \cos 3x)-1$
$2(\cos 14x + \cos 4x + \cos 11x + \cos 6x + 1+\cos 10x + \cos 8x + \cos 2 x) - 1$
However at the point I reached here then I got stuck it seems that this might not be the intended strategy. What could be done here to solve this problem?. Is there any trick or what?.
Could someone help me with an orderly method to solve this without much fuss?.
 A: If it is an identity, it will be true for any small angle $x$.
Using Taylor series, we have
$$\cos(nx)=1-\frac{n^2 }{2}x^2+O\left(x^4\right)$$ This makes the lhs to be
$$4 (1-32 x^2+1-18 x^2)(1 - 18 x^2+1-2 x^2)-1=15-560 x^2+O\left(x^4\right) \tag 1$$
Now, for the rhs
$$\sin(Kx)=K x-\frac{K^3 }{6}x^3+O\left(x^5\right)$$
$$\csc(x)=\frac{1}{x}+\frac{x}{6}+O\left(x^3\right)$$
$$\sin(Kx)\csc(x)=K+\frac{1}{6} \left(K-K^3\right) x^2+O\left(x^3\right)\tag 2$$
Compare $(1)$ and $(2)$. This gives $K=15$ and you can check that, for $K=15$
$$\frac{1}{6} \left(K-K^3\right)=-560$$
A: Let $x = \pi/4$. Then $\cos 2x = 0, \cos 6x = 0$, and $\cos 8x = 1$, which gives:
$$0 - 1 = \sin(K\pi/4) \cdot \csc \pi/4 \Rightarrow\sin(K\pi/4) = -\frac{1}{\sqrt2}$$
$$\Rightarrow K \pi/4 + 2\pi k= 5\pi/4, 7\pi/4$$
which only $K = 15$ satisfies from the given options.
A: Several completely different answers, and now here's another completely different answer, using something I just now learned about! There's a trig identity related to the Dirichlet kernel:
$$1+ 2\sum^n_{k=1} \cos(kx) = \frac{\sin \left((2n+1)\frac{x}{2}\right)}{\sin \frac{x}{2}} \tag{1}$$
I noticed you were getting lots of different multiples inside your cosines and then saw this formula and thought it might work. Here's what we're gonna do. First, let $z = 2x$, because it lets us get easier numbers on the left of the original, and it lets us make the RHS of $(1)$ equal to $\frac{\sin (n+1)z}{\sin z}$, which looks a lot like the RHS of your problem, doesn't it? We'll just say $\csc 30° = 2$ because it's a red herring (similar replacement on the RHS of $K = 2n + 1$). Now we have:
$$4(\cos 4z+\cos 3z)(\cos 3z + \cos z)-1= \frac{\sin ((2n+1)z)}{\sin z}$$
We're gonna head allllll the way back to basic algebra and FOIL this baby, dropping the RHS for a bit:
$$4(\cos 4z \cos 3z + \cos 4z \cos z + \cos 3z \cos z + \cos 3z \cos 3z)-1$$
Those product to sum formulas you were using before were on the right general track. Let's use them on this. They all need a $2$ out front, and conveniently we can grab one away from that $4$ on the outside, getting us:
$$2\left[(\cos 7z + \cos z) + (\cos 5z + \cos 3z) + (\cos 4z + \cos 2z) + (\cos 6z + \cos 0z)\right]-1= \\ 2(\cos 7z + \cos 6z + \cos 5z + \cos 4z + \cos 3z + \cos 2z + \cos z  +1)-1$$
At this point I think you can see where we're headed.  We can take care of the constants and we have:
$$1 + 2 \cos z + 2 \cos 2z + \cdots +2 \cos 7z= \frac{\sin (2n+1)z}{\sin z}$$
This means the $n$ on the sum in $(1)$ is $7$, and $K = 2n+1 = 15$.
And having done that, and seeing the other three answers, I cannot imagine any single one of them having been something I'd have been expected to know in pre-calculus. This is crazy difficult.
A: I show what $K$ must be. (I do not prove the identity.)
$\lim_{x\to 0} RHS=K.$
$\lim_{x\to 0} LHS=2\cdot 2(1+1)((1+1)-1=15.$
Therefore, $K$ is $15.$
