How to solve the following trigonometric equation?
$$\cos17x=20\cos x$$
I'm really awful in trigonometry. I tried division of both sides by $20$. Thanks.
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Are you sure about the equation? – egreg Sep 29 '14 at 11:02
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Well, $x=\pm \pi / 2$ obviously works, so it remains to check if this is unique $\pm 2n\pi$. – Macavity Sep 29 '14 at 11:04
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Hint: Use factor formula. – Hassan Muhammad Sep 29 '14 at 11:09
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1If all you need is the solution (without a formal proof), you can just graph the two functions and see where they meet: https://www.desmos.com/calculator/qb6n7onsig – Erel Segal-Halevi Sep 29 '14 at 11:21
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The function $f(x):=20\cos x-\cos(17x)$ is $2\pi$-periodic and even; therefore it suffices to analyze what happens in the $x$-interval $[0,\pi]$. Write $x:={\pi\over2}-y$ and consider the function $$g(y):=f\left({\pi\over2}-y\right)=20\sin y-\sin(17 y)\qquad\left(-{\pi\over2}\leq y\leq{\pi\over2}\right)$$ instead. Since $g$ is odd we can even restrict to $0\leq y\leq{\pi\over2}$.
When $0<y\leq{1\over10}$ truncating the $\sin$ series after a negative term provides the estimate $$\sin y>y-{y^3\over6}= y\left(1-{y^2\over6}\right)>{9\over10}y\ .$$ It follows that $$20\sin y-\sin(17y)> 18 y-17 y>0\qquad\left(0<y\leq{1\over10}\right)\ .$$ Furthermore one has $$20\sin y-\sin(17y)\geq 20\sin{1\over10}-1>20{9\over10}{1\over10}-1={4\over5}>0\qquad\left({1\over10}\leq y\leq{\pi\over2}\right)\ .$$ It follows that the only zero of $g$ is at $y=0$, which corresponds to $x={\pi\over2}$. Therefore the solutions of the original equation are the odd multiples of ${\pi\over2}$.
Christian Blatter
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Thanks. Could you please explain how to prove that thing that comes from alternate series? – MelanieJohansson Sep 29 '14 at 12:46
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Hint: The key to a simple solution seems to me to be
First to identify where $-1\le 20\cos x \le 1$ because this is necessary for a solution.
Then show that the curve of $20\cos x$ is steeper than that of $\cos 17 x$ in those intervals (e.g. use the derivative.)
Then show that the solutions where both expressions are simultaneously zero are the only ones (and pin these down)
Here a very rough sketch shows what is going on, and avoids a lot of unnecessary algebra.
Mark Bennet
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There probably is a simpler solution, but ... you can use Chebyshev polynomials : if $c=\cos(x)$, your equation becomes $T_{17}(c)=20c$, or
$$ c(65536c^{16} - 278528c^{14} + 487424c^{12} - 452608c^{10} + 239360c^8 - 71808c^6 + 11424c^4 - 816c^2 - 3)=0 $$
But the polynomial $A=65536c^{16} - 278528c^{14} + 487424c^{12} - 452608c^{10} + 239360c^8 - 71808c^6 + 11424c^4 - 816c^2 - 3$ has only two opposite roots, both of absolute value $>1$. So $c=0$, and $x=\pm \frac{\pi}{2}+2n\pi$ as Macavity guessed.
Ewan Delanoy
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