Solve $\cos^n x + \sin^n x =1 $

Question

the solutions of this equation as a function of the value of $n$??

\begin{align} \cos^n x + \sin^n x =1 \end{align}

I already found the solution if n is odd,

Answer 1

Hint:

For positive $n$: You know that for any $x$, $$\cos^2 x + \sin^2 x = 1,$$ and both the terms on the left are positive. For $n = 2k$, if $x$ is a solution, then you have $\cos^{2k} x + \sin^{2k} x = 1$, so $$(\cos^2 x)^k + (\sin^2 x)^k = 1$$.

Look at those two displayed equations and ask yourself, "How are $\cos^2 x $ and $(\cos^2 x)^k$ related? Which is larger in general?"

For negative $n$: A similar argument should work, but with the inequality reversed.

Answer 2

If $n$ is odd, we can write it as $n = 2k+1, k \ge 1$. $$1 = |\sin^{2k+1}x + \cos^{2k+1}x| \le |\sin x|\cdot \sin^{2k}x + |\cos x|\cdot \cos^{2k}x \le \sin^{2k}x+\cos^{2k}x \le \sin^2 x + \cos^2x = 1.$$ Thus, $|\sin x| = 1, 0$, and you can find $x$ from this.

If $n = 1$, then $$\sin x + \cos x = 1\implies (\sin x+\cos x)^2 = 1 \implies \sin (2x) = 0 \implies 2x = m\pi \implies x = \dfrac{m\pi}{2}, m \in \mathbb{Z}.$$

If $n$ is even, then $$n \ge 2 \implies 1 = \sin^2x + \cos^2 x \ge \sin^n x+\cos^n x = 1\implies \cos^2 x = 1, 0 \implies \cos x = 0, \pm 1 \implies x = m\pi, \pm\dfrac{\pi}{2} + 2m\pi, m \in \mathbb{Z}.$$

Answer 3

There is a nice geometrical interpretation if you will.

We know $\cos^2 x + \sin^2 x = 1$ for all $x$. You want to solve $\cos^n x + \sin^n x = 1$. If $a = \cos x$ and $b = \sin x$, then you want to solve the simultaneous equations: $$ a^2 + b^2 = 1,\quad\quad\mbox{circunference}\\ a^n + b^n = 1,\quad\mbox{super-circunference} $$

The circunference and super-circunference for $n > 2$ will intercept only in the points $(1, 0)$, $(0, 1)$, $(-1, 0)$, $(0, -1)$. The greater than $n$, the more the super-circunference will look like a square.

Answer 4

If $\cos x\sin x\not=0$, then $|\cos x|=\sqrt{1-\sin^2x}$ and $|\sin x|=\sqrt{1-\cos^2x}$ are both strictly less than $1$, which implies $\cos^nx+\sin^nx\lt\cos^2x+\sin^2x=1$ for $n\gt2$. For $n=1$, $\cos x+\sin x=1$ implies $\cos^2x+2\cos x\sin x+\sin^2x=1$, which implies $\cos x\sin x=0$. In sum, when $n$ is a positive integer other than $2$, $\cos^nx+\sin^nx=1$ implies $x$ is a multiple of $\pi/2$. If $n$ is even, any (integer) multiple of $\pi/2$ is a solution. If $n$ is odd, the multiple must be congruent to $0$ or $1$ mod $4$.

For negative integers $n$, we cannot have $\cos x\sin x=0$. If $n\lt0$ is even, then $\cos x\sin x\not=0$ implies $\cos^n+\sin^n\gt\cos^2+\sin^2=1$, so the equation has no solutions. If $n\lt0$ is odd, however, there are solutions. In particular, there is always a solution with $-\pi/2\lt x\lt 0$, since $\sec x\to\infty$ as $x\to-\pi/2^+$ while $\csc x\to-\infty$ as $x\to0^-$.

Finally, if $n=0$, the equation presumably has no solutions, since $\cos^0x+\sin^0x=1+1=2\not=1$ if $\cos x\sin x\not=0$ and isn't clearly defined if $\cos x\sin x=0$.

Answer 5

For $n=1$ it can be solved by squaring both sides and OP also solved. For $n=2$ the equation is true for all $x$. So assume that $n\geq 3$.

Let $f(x)=\cos^n x+\sin^n x$. Then $f'(x)=n\sin x\cos x(\sin^{n-2}x-\cos^{n-2} x)$. For extrema, $f'(x)=0$ gives $\cos x=0$, $\sin x=0$, $\tan x=1$ or $\tan x=-1$ for $n$ even. But for $\tan x=\pm1$, the value of $f(x)$ can be at most $(\frac{1}{\sqrt2})^n+(\frac{1}{\sqrt2})^n=2^{1-\frac n2}<1$ since $1-\frac n2<0$ due to the assumption. And we are done.

Answer 6

Firstly, let’s assume that both $\cos x$ and $\sin x$ are non-zero, then $$ \cos ^n x<\cos ^2 x \textrm{ and } \sin ^n x<\sin ^2 x. $$ Adding them together, we get $$ \sin ^n x+\cos ^n x<\sin ^2 x+\cos ^2 x=1 \Rightarrow \sin ^n x+\cos ^n x=1 \textrm{ has no solution.} $$ Therefore $\cos x=0$ or $\sin x=0.$ Now we need to consider the parity of $n.$

A. When $n$ is even,

(i) $\cos x=0 \wedge \sin x= \pm 1 \Rightarrow x=n \pi+\frac{\pi}{2}$

(ii) $\sin x=0 \wedge \cos x= \pm 1 \Rightarrow x=n \pi$ and therefore

$$\boxed{x=n \pi+\frac{\pi}{2} \textrm{ or }n \pi \textrm{ for any }n \in \mathbb{Z}.}$$

B. When $n$ is odd,

(i) $\cos x=0 \wedge \sin x=1 \Rightarrow x=2 n \pi+\frac{\pi}{2}$

(ii) $\sin x=0 \wedge \cos x=1 \Rightarrow x=2n \pi$ and therefore

$$\boxed{x=2 n \pi+\frac{\pi}{2} \textrm{ or }2 n \pi \textrm{ for any }n \in \mathbb{Z}.}$$

Answer 7

Note that $\cos^n x + \sin^n x =1 $ has solution if and only if both terms are not negative, therefore we need to consider just two cases with $n$ even

• $0<|\cos x|<1$ and $n>2$

$$\cos^n x + \sin^n x < \cos^2 x+\sin^2 x=1$$

• $|\cos x|=0\;\lor\; |\cos x|=1$

$$\cos^n x + \sin^n x =1+0=1$$