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Solve the equation $\cos^n(x) - \sin^n(x)=1,n \in \mathbb{N}-\{0\}$

If $n$ is even then $\cos^n(x) = \sin^n(x)+1$ is only possible if $\sin(x)=0$ therefore the solution is $x=k\pi, k \in \mathbb{Z}$.

I'm having problems with $n$ odd case.

UPDATE

For $n=1$ we have $\cos(x) - \sin(x)=1$ and by squaring we get $sin(x)cos(x)=0$ which leads to $x=k\pi$ or $x=\pm \frac \pi 2 + k\pi$. From these solutions, only $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are valid.

Also $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are solutions for all odd $n$

  • For the odd case, $x=2k\pi$ seems obvious (I hope). There is another one which is simple. – Claude Leibovici Sep 25 '16 at 06:41
  • @ClaudeLeibovici $x=2k\pi$ is working for all n, regardless the parity –  Sep 25 '16 at 06:46
  • If you plot the function s.t. $n>1$ and $n$ odd, you'll see there are such "trivial" solutions only. On the other hand, you also have "non-trivial" solutions if $n=1$ (i.e. $x=2\pi k-\pi/2$). This happens, since $cos$ and $sin$ "shrink" rapidly as $n$ gets larger. – Math.StackExchange Sep 25 '16 at 06:48
  • @Math.StackExchange Good to know, so the $n=1$ case should be separately investigated. –  Sep 25 '16 at 06:51
  • For sure ! Did you think about $(2k-\frac 12)\pi$ ? – Claude Leibovici Sep 25 '16 at 06:55
  • That's what I got from Wolfram Alpha. For $n=1$, you can use the formula $\cos(x)=\cos^2(x/2)-\sin^2(x/2)$. – Math.StackExchange Sep 25 '16 at 06:57
  • As mentioned below, for $n$ odd the solutions are $2\pi k$ and $2\pi k-\pi/2$. This can be proved by checking the derivative of $\sin^n(x)-\cos^n(x)$ for $n>1$ odd. – Math.StackExchange Sep 25 '16 at 07:07

3 Answers3

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It suffices to investigate the solutions over $[0,2\pi)$, as all solutions will be some integer multiple of $2\pi$ away from a solution in $[0,2\pi)$. Let $f(x) = \cos^n(x)-\sin^n(x)$, with $n$ odd and $n\ge 3$. Then $$f'(x) = n(\cos^{n-1}(x)(-\sin x) - \sin^{n-1}(x)\cos(x)) = -n\sin(x)\cos(x)(\cos^{n-2}(x)+\sin^{n-2}(x)) $$ It can easily be seen that $f'$ has zeros at $0$, $\frac{\pi}{2}$, $\frac{3\pi}{4}$, $\pi$, $\frac{3\pi}{2}$, and $\frac{7\pi}{4}$, with $f'<0$ on $(0,\frac{\pi}{2})\cup(\frac{3\pi}{4},\pi)\cup(\frac{3\pi}{2},\frac{7\pi}{4})$ and $f'>0$ on $(\frac{\pi}{2},\frac{3\pi}{4})\cup(\pi,\frac{3\pi}{2})\cup(\frac{7\pi}{4},2\pi)$.

Now \begin{align} f(0) &= 1 \\ f(\frac{\pi}{2}) &= -1 \\ f(\frac{3\pi}{4}) &= -2\left(\frac{1}{\sqrt{2}}\right)^n \\ f(\pi) &= -1 \\ f(\frac{3\pi}{2}) &= 1 \\ f(\frac{7\pi}{4}) &= 2\left(\frac{1}{\sqrt{2}}\right)^n \end{align} From this, we see that $0$ and $\frac{3\pi}{2}$ are solutions. Furthermore, since $f'<0$ on $(0,\frac{\pi}{2})$, we have that $f(x) < f(0) = 1$ for $x\in(0,\frac{\pi}{2})$. Similarly, since $f'>0$ on $(\frac{\pi}{2},\frac{3\pi}{4})$, we have that $f(x)<f(\frac{3\pi}{4}) = -2\left(\frac{1}{\sqrt{2}}\right)^n < 1 $ for $x\in(\frac{\pi}{2},\frac{3\pi}{4})$. Applying similar reasoning to the other four intervals, we find that $f(x)<1$ for all $x\in[0,2\pi)$ except $x=0$ and $x=\frac{3\pi}{2}$.

It follows that the only solutions in $[0,2\pi)$ are $x=0$ and $x=\frac{3\pi}{2}$, so the only solutions in $\mathbb{R}$ are $\boxed{x = 2k\pi}$ and $\boxed{x = \frac{3\pi}{2} + 2k\pi}$ for $k\in\mathbb{Z}$.

Joey Zou
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4

I found a simple proof. For $n \ge 3$ odd we have $\cos^n(x) + (-\sin(x))^n=1 \tag1$ First, we know $\sin^2(x) + \cos^2(x)=1$

Also $|\cos^n(x)|\lt \cos^2(x)$ and $|\sin^n(x)|\lt \sin^2(x)$ for $x \not = k\pi$ and $x \not = \pm \frac \pi 2 + k\pi$ therefore the equality (1) cannot hold for $x \not = k\pi$ and $x \not = \pm \frac \pi 2 + k\pi$

It's easy to conclude from here.

  • +1, brilliant answer. More succinctly, $|\cos^n(x) - \sin^n(x)| \le |\cos^n(x)| + |\sin^n(x)| < |\cos^2(x)| + |\sin^2(x)| = \cos^2(x) + \sin^2(x) = 1$. – Caleb Stanford Sep 25 '16 at 07:50
  • @6005 Right, it is better the way you put it. –  Sep 25 '16 at 07:57
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It will only satisfy at 2kπ where, k is an integer. Because when n is odd then at only even multiples of π theq equation will be satisfied.

Seemant Shekhar
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