IMO 1961 Problem 3
We seek all real numbers $x$ satisfying
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 39m28s
Problem
Solve the equation
$\cos^n{x} - \sin^n{x} = 1$
where n is a given positive integer.
Exploration
We seek all real numbers $x$ satisfying
$$\cos^n x-\sin^n x=1,$$
where $n\in\mathbb N$ is fixed.
Since $\sin x,\cos x\in[-1,1]$, the left-hand side is bounded. Equality with $1$ suggests an extremal phenomenon. Writing
$$\cos^n x=1+\sin^n x$$
immediately gives restrictions, because $\cos^n x\le 1$.
The first five values of $n$ suggest the pattern.
For $n=1$,
$$\cos x-\sin x=1.$$
Testing $x=0$ gives a solution. Testing $x=\tfrac{3\pi}{2}$ gives
$$0-(-1)=1,$$
also a solution.
For $n=2$,
$$\cos^2x-\sin^2x=1$$
becomes
$$\cos 2x=1,$$
hence $x=k\pi$.
For $n=3$,
$$\cos^3x-\sin^3x=1.$$
Again $x=0$ works, and $x=\tfrac{3\pi}{2}$ gives
$$0-(-1)^3=1.$$
No other obvious solutions appear.
For $n=4$,
$$\cos^4x-\sin^4x = (\cos^2x-\sin^2x)(\cos^2x+\sin^2x) = \cos 2x,$$
so again $x=k\pi$.
For $n=5$, the same behavior as $n=3$ appears.
This suggests a parity distinction. If $n$ is even, both powers are nonnegative, so
$$\cos^n x\le 1, \qquad \sin^n x\ge 0.$$
To obtain equality
$$\cos^n x-\sin^n x=1,$$
we would need simultaneously
$$\cos^n x=1, \qquad \sin^n x=0.$$
That forces $\cos x=\pm1$ and $\sin x=0$, hence $x=k\pi$.
If $n$ is odd, then $\sin^n x$ may be negative. Since
$$\cos^n x\le 1,$$
the equation
$$\cos^n x=1+\sin^n x$$
requires
$$1+\sin^n x\le 1,$$
hence $\sin^n x\le 0$. Equality in the upper bound $\cos^n x\le 1$ again looks necessary. Since the right-hand side is at most $1$, we must have
$$\cos^n x=1.$$
Then $\sin^n x=0$, unless $\sin^n x=-1$ and $\cos^n x=0$, which also gives $1$. For odd $n$, $\sin x=-1$ is possible.
The delicate point is proving rigorously that no intermediate values can occur. A careless argument might overlook the case
$$\cos^n x=0,\qquad \sin^n x=-1,$$
when $n$ is odd.
Problem Understanding
We are asked to determine all real numbers $x$ satisfying
$$\cos^n x-\sin^n x=1,$$
for a fixed positive integer $n$.
This is a Type A problem, a classification problem. We must find every solution and prove two facts: first, that every claimed solution satisfies the equation; second, that no other real number satisfies it.
The objects involved are real trigonometric values raised to an integer power. The difficulty lies in handling all positive integers $n$ simultaneously, because parity changes the behavior of powers. When $n$ is even, both terms are nonnegative. When $n$ is odd, negative values of $\sin x$ remain negative after exponentiation, creating an additional possible mechanism for equality.
The expected answer is:
$$x=k\pi \quad\text{for all integers }k,$$
when $n$ is even, and
$$x=2k\pi \quad\text{or}\quad x=\frac{3\pi}{2}+2k\pi$$
for all integers $k$, when $n$ is odd.
The intuition is that the quantity
$$\cos^n x-\sin^n x$$
can equal $1$ only in extremal situations, because $\cos^n x\le 1$. Equality forces one of the terms to attain an endpoint value.
Proof Architecture
We will prove the result by separating the parity cases for $n$.
Claim 1. If $n$ is even, every solution satisfies
$$\cos^n x=1 \quad\text{and}\quad \sin^n x=0.$$
This follows from the inequalities
$$0\le \cos^n x\le 1, \qquad \sin^n x\ge 0,$$
whose sum can equal $1$ only at the endpoint.
Claim 2. If $n$ is even, the complete solution set is
$$x=k\pi, \qquad k\in\mathbb Z.$$
We solve the conditions from Claim 1 explicitly and verify them.
Claim 3. If $n$ is odd, every solution must satisfy one of the two systems
$$\cos x=1,\qquad \sin x=0,$$
or
$$\cos x=0,\qquad \sin x=-1.$$
This uses the bounds
$$-1\le \sin^n x\le 1, \qquad \cos^n x\le 1,$$
and the identity
$$\cos^n x=1+\sin^n x.$$
Claim 4. If $n$ is odd, the complete solution set is
$$x=2k\pi \quad\text{or}\quad x=\frac{3\pi}{2}+2k\pi, \qquad k\in\mathbb Z.$$
We solve the systems from Claim 3 and check them directly.
The hardest direction is excluding all other values of $x$ in the odd case, because one must avoid overlooking the possibility $\sin^n x=-1$. The most fragile step is the extremal argument showing that $\cos^n x$ must equal either $1$ or $0$.
Solution
We distinguish two cases according to the parity of $n$.
Case 1: $n$ is even
Since $n$ is even,
$$0\le \cos^n x\le 1$$
and
$$\sin^n x\ge 0$$
for every real number $x$.
The equation
$$\cos^n x-\sin^n x=1$$
may be rewritten as
$$\cos^n x=1+\sin^n x.$$
Because $\sin^n x\ge 0$, the right-hand side satisfies
$$1+\sin^n x\ge 1.$$
Because $\cos^n x\le 1$, equality is possible only if
$$\cos^n x=1 \quad\text{and}\quad 1+\sin^n x=1.$$
Hence
$$\sin^n x=0.$$
Since $n$ is positive,
$$\sin^n x=0 \iff \sin x=0.$$
Also,
$$\cos^n x=1$$
with even $n$ implies
$$\cos x=\pm1.$$
The condition $\sin x=0$ already yields
$$x=k\pi, \qquad k\in\mathbb Z,$$
and for such $x$ we indeed have
$$\cos x=\pm1.$$
We verify these candidates:
if $x=k\pi$, then
$$\sin x=0, \qquad \cos x=(-1)^k.$$
Since $n$ is even,
$$\cos^n x=1, \qquad \sin^n x=0,$$
and thus
$$\cos^n x-\sin^n x=1.$$
This establishes that for even $n$ the complete set of solutions is
$$x=k\pi, \qquad k\in\mathbb Z.$$
Certification: this step proves both necessity and sufficiency for even $n$; merely solving $\cos^n x=1$ without proving $\sin^n x=0$ would leave logical gaps.
Case 2: $n$ is odd
Since $n$ is odd,
$$-1\le \sin^n x\le 1, \qquad -1\le \cos^n x\le 1.$$
The equation becomes
$$\cos^n x=1+\sin^n x.$$
Because $\sin^n x\le 1$,
$$1+\sin^n x\le 2.$$
More importantly, since $\cos^n x\le 1$,
$$1+\sin^n x\le 1,$$
which implies
$$\sin^n x\le 0.$$
Now
$$\cos^n x-\sin^n x=1$$
with
$$\cos^n x\le 1$$
and
$$-\sin^n x\le 1$$
shows that the left-hand side is the sum of two quantities, each at most $1$. Equality can occur only in one of the following extremal ways:
Either
$$\cos^n x=1 \quad\text{and}\quad -\sin^n x=0,$$
which gives
$$\cos x=1, \qquad \sin x=0,$$
or
$$\cos^n x=0 \quad\text{and}\quad -\sin^n x=1,$$
which gives
$$\cos x=0, \qquad \sin x=-1.$$
Solving these systems yields:
From
$$\cos x=1, \qquad \sin x=0,$$
we obtain
$$x=2k\pi, \qquad k\in\mathbb Z.$$
From
$$\cos x=0, \qquad \sin x=-1,$$
we obtain
$$x=\frac{3\pi}{2}+2k\pi, \qquad k\in\mathbb Z.$$
We verify them directly.
If
$$x=2k\pi,$$
then
$$\cos x=1, \qquad \sin x=0,$$
hence
$$\cos^n x-\sin^n x=1.$$
If
$$x=\frac{3\pi}{2}+2k\pi,$$
then
$$\cos x=0, \qquad \sin x=-1.$$
Since $n$ is odd,
$$\sin^n x=-1,$$
so
$$\cos^n x-\sin^n x = 0-(-1) = 1.$$
Certification: this step excludes every non-extremal possibility; overlooking the case $\sin x=-1$ would incorrectly omit valid solutions for odd $n$.
Hence the complete solution set is
$$\boxed{ \begin{cases} x=k\pi, & \text{if } n \text{ is even},\[6pt] x=2k\pi \text{ or } x=\dfrac{3\pi}{2}+2k\pi, & \text{if } n \text{ is odd}, \end{cases} \qquad k\in\mathbb Z. }$$
Verification of Key Steps
The first delicate step is the even case. Starting from
$$\cos^n x-\sin^n x=1,$$
with even $n$,
$$0\le \cos^n x\le 1, \qquad \sin^n x\ge 0.$$
Thus
$$1=\cos^n x-\sin^n x\le \cos^n x\le 1.$$
Both inequalities are equalities, so
$$\cos^n x=1 \quad\text{and}\quad \sin^n x=0.$$
A careless argument might stop after $\cos^n x=1$ and forget that $\sin^n x=0$ is also forced.
The second delicate step is the odd case. Rewrite the equation as
$$1=\cos^n x+(-\sin^n x).$$
Since
$$\cos^n x\le 1$$
and
$$-\sin^n x\le 1,$$
the sum equals $1$. If
$$-\sin^n x<0,$$
then
$$\cos^n x>1,$$
impossible. Hence
$$-\sin^n x\in[0,1].$$
Writing
$$1=\cos^n x+(-\sin^n x)$$
with both summands in $[0,1]$, the only possibilities are
$$(1,0) \quad\text{or}\quad (0,1),$$
because any interior pair would force one coordinate to exceed $1$. A careless proof might miss the second endpoint pair.
The third delicate step is converting powers back to trigonometric values. For odd $n$,
$$a^n=1 \iff a=1, \qquad a^n=-1 \iff a=-1,$$
for real $a\in[-1,1]$. For even $n$,
$$a^n=1 \iff a=\pm1.$$
Ignoring parity here would produce incorrect solution sets.
Alternative Approaches
A different strategy uses the identity
$$\cos^2x+\sin^2x=1$$
and introduces
$$u=\cos^n x, \qquad v=\sin^n x.$$
For even $n$, one has
$$u,v\ge 0,$$
and
$$u-v=1$$
forces
$$u=1,\quad v=0.$$
For odd $n$, one has
$$u\le 1,\qquad -v\le 1,$$
and the equation becomes
$$u+(-v)=1.$$
Bounding both summands by $1$ yields the same extremal analysis.
Another possible approach separates the cases according to the sign of $\sin x$. If $\sin x\ge 0$, then
$$\cos^n x-\sin^n x\le \cos^n x\le 1,$$
and equality forces
$$\cos^n x=1,\qquad \sin x=0.$$
If $\sin x<0$, then only odd $n$ can produce additional solutions, since even powers remove the sign. This route is shorter but less systematic, because parity enters midway rather than from the beginning.