Kvant Math Problem 922
Let
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Solve time: 3m30s
Source on kvant.digital
Problem
Prove that the equation $$\sin^px+\cos^qx=1,$$, where $p$ and $q$ are positive numbers, has a solution on the interval $0\lt x\lt \dfrac{\pi}{2}$ if and only if $(p-2)(q-2)\lt0$ or $p=q=2$.
A. M. Sedletskii
Exploration
Let
$f(x)=\sin^p x+\cos^q x,\qquad x\in(0,\tfrac{\pi}{2}),\quad p,q>0.$
At the endpoints,
$f(0)=0^p+1^q=1,\qquad f!\left(\tfrac{\pi}{2}\right)=1^p+0^q=1,$
so both boundary points satisfy the equation, although they are excluded.
Inside the interval, both $\sin x$ and $\cos x$ lie in $(0,1)$, so increasing or decreasing the exponents changes each term monotonically. The decisive comparison is between exponents $2$ and the behavior of $t^p$ relative to $t^2$ on $(0,1)$.
If $p,q\ge2$, then $t^p\le t^2$ and $t^q\le t^2$, suggesting $f(x)\le1$ with equality only at the endpoints, hence no interior solution unless $p=q=2$ makes $f\equiv1$.
If $p,q\le2$, then both inequalities reverse, suggesting $f(x)\ge1$ with equality again only at endpoints, so no interior solution.
The remaining mixed cases $(p<2<q)$ and $(q<2<p)$ should force a sign change of $f(x)-1$ near $0$ and near $\tfrac{\pi}{2}$. The critical mechanism is the competition between $x^p$ and $x^2$ as $x\to0$, and similarly near $\tfrac{\pi}{2}$ via symmetry.
The key point is therefore the local expansion near the endpoints and comparison of powers.
Problem Understanding
This is a Type A problem: characterize all positive pairs $(p,q)$ for which
$\sin^p x+\cos^q x=1$
has a solution in $(0,\tfrac{\pi}{2})$.
We must show that an interior solution exists if and only if either $(p-2)(q-2)<0$ or $p=q=2$.
The core difficulty is determining whether the function $f(x)$ can cross the level $1$ inside the interval, given that it equals $1$ at both endpoints.
Proof Architecture
First, we establish the comparison principle: for $t\in(0,1)$, $t^r\le t^2$ when $r\ge2$, and $t^r\ge t^2$ when $0<r\le2$, with strictness unless $t\in{0,1}$.
Second, we deduce that if $p,q\ge2$, then $f(x)\le1$ on $(0,\tfrac{\pi}{2})$, and equality holds only when $p=q=2$.
Third, we deduce that if $p,q\le2$, then $f(x)\ge1$ on $(0,\tfrac{\pi}{2})$, with equality only at endpoints.
Fourth, we analyze asymptotics near $0$: we compare $\sin^p x$ with $x^p$ and $\cos^q x$ with $1-\frac{q}{2}x^2$ to determine the sign of $f(x)-1$ depending on whether $p<2$ or $p>2$.
Fifth, we repeat the symmetric analysis near $\tfrac{\pi}{2}$.
Finally, we combine endpoint sign information with continuity to force an interior root exactly in the mixed cases.
The most delicate point is the asymptotic comparison near the endpoints, where the relative sizes of $x^p$ and $x^2$ determine the sign.
Solution
Let $x\in(0,\tfrac{\pi}{2})$ and define
$f(x)=\sin^p x+\cos^q x.$
For $t\in(0,1)$ the function $r\mapsto t^r$ is strictly decreasing, hence if $r\ge2$ then $t^r\le t^2$, while if $0<r\le2$ then $t^r\ge t^2$.
Assume first that $p\ge2$ and $q\ge2$. Then for all $x\in(0,\tfrac{\pi}{2})$,
$\sin^p x\le \sin^2 x,\qquad \cos^q x\le \cos^2 x,$
hence
$f(x)\le \sin^2 x+\cos^2 x=1.$
Equality in both inequalities holds only if $\sin x\in{0,1}$ and $\cos x\in{0,1}$ simultaneously, which is impossible for $x\in(0,\tfrac{\pi}{2})$. Therefore $f(x)<1$ in the open interval unless both exponents equal $2$, in which case $f(x)\equiv1$ for all $x$. In the latter case every interior point is a solution.
Assume next that $p\le2$ and $q\le2$. Then
$\sin^p x\ge \sin^2 x,\qquad \cos^q x\ge \cos^2 x,$
so
$f(x)\ge \sin^2 x+\cos^2 x=1.$
Again equality inside the interval is impossible unless both exponents equal $2$.
Hence if $(p-2)(q-2)\ge0$ and $(p,q)\ne(2,2)$, the function satisfies either $f(x)<1$ or $f(x)>1$ for all interior $x$, so no solution exists in $(0,\tfrac{\pi}{2})$, while if $p=q=2$ every point is a solution.
Now assume $(p-2)(q-2)<0$. Without loss of generality let $p<2<q$. The case $p>2>q$ follows symmetrically.
Near $x=0$, we use standard bounds $\sin x\sim x$ and $\cos x=1-\frac{x^2}{2}+o(x^2)$. More precisely, there exist constants $c_1,c_2>0$ such that for sufficiently small $x$,
$c_1 x \le \sin x \le x,\qquad 1-\frac{x^2}{2}\le \cos x \le 1.$
Then
$\sin^p x \ge c_1^p x^p.$
Also, using $(1-y)^q \le 1-qy$ for small $y>0$,
$\cos^q x \ge \left(1-\frac{x^2}{2}\right)^q = 1-\frac{q}{2}x^2+o(x^2),$
so for small $x$ there exists $C>0$ such that
$\cos^q x \ge 1-Cx^2.$
Therefore
$f(x)-1 \ge c_1^p x^p - Cx^2.$
Since $p<2$, the term $x^p$ dominates $x^2$ as $x\to0$, hence $f(x)>1$ for all sufficiently small $x>0$.
Now consider $x$ near $\tfrac{\pi}{2}$. Write $x=\tfrac{\pi}{2}-t$ with $t\to0^+$. Then $\cos x=\sin t$ and $\sin x=\cos t$. The same argument applied with exponents swapped yields
$f(x)-1 \approx t^q - Ct^2$
with $q>2$, hence $t^q$ is dominated by $t^2$ and so $f(x)<1$ for sufficiently small $t>0$.
Thus there exist points $x_1,x_2\in(0,\tfrac{\pi}{2})$ such that $f(x_1)>1$ and $f(x_2)<1$. By continuity of $f$, there exists $x\in(x_1,x_2)$ such that $f(x)=1$.
The case $p>2>q$ is symmetric, yielding the same conclusion.
Hence an interior solution exists exactly when $(p-2)(q-2)<0$ or $p=q=2$, which is the required characterization.
Verification of Key Steps
The crucial endpoint analysis uses the comparison between $x^p$ and $x^2$. For $p<2$, the ratio $x^p/x^2=x^{p-2}$ diverges to $+\infty$ as $x\to0^+$, ensuring that the positive contribution from $\sin^p x$ dominates the negative quadratic deviation of $\cos^q x$ from $1$. This guarantees $f(x)>1$ near $0$.
For $p>2$, the same ratio tends to $0$, so the quadratic deficit from $\cos^q x$ dominates, forcing $f(x)<1$ near $0$.
The symmetric argument near $\tfrac{\pi}{2}$ replaces $x$ by $t=\tfrac{\pi}{2}-x$ and exchanges the roles of $p$ and $q$, producing the opposite inequality when the exponent at that endpoint exceeds $2$.
These two endpoint behaviors are incompatible unless $(p-2)(q-2)<0$, ensuring a sign change inside the interval and therefore an interior root.
Alternative Approaches
A different method replaces asymptotic estimates by the convexity of $t\mapsto t^{\alpha}$ on $(0,1)$ and a careful study of the maximum of $\sin^p x+\cos^q x$ via differentiation. One can show that the unique interior critical point occurs where $p\sin^{p-2}x\cos^2 x=q\cos^{q-2}x\sin^2 x$, and then analyze whether this critical value exceeds $1$. The endpoint-asymptotic approach is shorter and avoids solving transcendental critical point equations.