1
$\begingroup$

If $x=\frac{\pi}{34}$, then find the value of $$S=4\sin (x)\,\big(\sin(3x)+\sin (7x)+\sin(11x)+\sin(15x)\big).$$

Source: Joszef Wildt International Math Competition Problem

My attempt:

$$\sin(3x)+\sin(15x)=2\sin(9x)\cos(6x)$$

$$\sin(7x)+\sin(11x)=2\sin(9x)\cos(2x)$$

So $$S=8\sin(x)\sin(9x)\big(\cos(6x)+\cos(2x)\big)$$

$$S=16\sin(x)\sin(9x)\cos(4x)\cos(2x)$$ $$S=16\sin(x)\sin(9x)\sin(13x)\sin(15x)$$

any clue here?

$\endgroup$
2
  • $\begingroup$ I fixed the first expression which was not clear (at least to me !). Cheers $\endgroup$ Commented May 12, 2020 at 13:20
  • 5
    $\begingroup$ Hint: Your last expression is equal to $$S=16\cos\frac{\pi}{17}\cos\frac{2\pi}{17}\cos\frac{4\pi}{17}\cos\frac{8\pi}{17}.$$ $\endgroup$ Commented May 12, 2020 at 13:26

3 Answers 3

3
$\begingroup$

Alternatively, telescope the expression with

\begin{align} &4\sin x(\sin 3x+\sin 7x+\sin 11x+\sin 15x)\\ =&\frac{2\sin2x}{\cos x}(\sin 3x+\sin 7x+\sin 11x+\sin 15x)\\ =&\frac 1{\cos x} [ (\cos x-\cos5x) + (\cos 5x-\cos9x) + (\cos 9x-\cos13x) + (\cos 13x-\cos17x)]\\ =& \frac{1}{\cos x}(\cos x - \cos 17x) =1-\frac{\cos\frac\pi2}{\cos\frac{\pi}{34}}=1 \end{align}

$\endgroup$
2
$\begingroup$

as You simplified and as @WE Tutorial School pointed out, $$S=16\cos\frac{\pi}{17}\cos\frac{2\pi}{17}\cos\frac{4\pi}{17}\cos\frac{8\pi}{17}.$$

Applying $\sin 2x = 2 \sin x \cos x$ 4 times,

$$ S \cdot \sin \frac \pi{17} = \sin \frac {16\pi}{17}$$

$$ S = \frac {\sin \left( \pi - \frac \pi{17} \right)}{\sin \frac \pi{17}}$$

$$S=1$$

$\endgroup$
1
$\begingroup$

As you calculated, we have

$$S=16\sin(x)\sin(9x)\cos(4x)\cos(2x).$$

Using $\sin x=\cos(\tfrac{\pi}{2}-x)$, we get

$$S\left(\frac{\pi}{34}\right)=16\cos\frac{\pi}{17}\cos\frac{2\pi}{17}\cos\frac{4\pi}{17}\cos\frac{8\pi}{17}.$$

Let $\zeta=e^{2\pi i/17}$. Noting that $\zeta^k+\zeta^{-k}=2\cos \frac{2k\pi}{17}$ and $\cos \frac{\pi}{17}=-\cos \frac{16\pi}{17}$, we have

$$S\left(\dfrac{\pi}{34}\right)=-(\zeta+\zeta^{-1})(\zeta^2+\zeta^{-2})(\zeta^4+\zeta^{-4})(\zeta^8+\zeta^{-8})=\boxed{1}$$

using $\sum_{k=1}^{16} \zeta^k=-1$.

$\endgroup$
2
  • $\begingroup$ But $\cos(4x)=\sin(13x)$ is correct as $4x+13x=\frac{\pi}{2}$ why its' wrong? $\endgroup$ Commented May 12, 2020 at 15:30
  • $\begingroup$ @UmeshShankar correct, I edited ;) $\endgroup$
    – rae306
    Commented May 12, 2020 at 15:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .