Functional equation (is solution unique) 
*

*Let $f:[0,1]\rightarrow \mathbb{R}$ - cont. diff. function. Is it true that equation $\cos{t}f(\sin{t})+\sin{t}f(\cos{t})=1, t \in [0,\pi/2]$ has only solution $f(x)=\sqrt{1-x^2}$? How can we prove the uniqueness?

*Same question with $f$ just continuous.
 A: ANALYSIS
Note that the related equation
$$
\cos(t)\cdot x+\sin(t)\cdot y=1
$$
defines a  tangent to the unit circle through the point $P=(\cos(t),\sin(t))$. Points on this tangent can be described by the vector function with coordinates
$$
\begin{align}
x(s)&=\cos(t)-s\cdot\sin(t)\\
y(s)&=\sin(t)+s\cdot\cos(t)
\end{align}
$$
Furthermore, we must arrange things so that $x(s)=f(\sin(t))$ and $y(s)=f(\cos(t))$ in order to match this related equation and the original problem. Therefore $f$ satisfies
$$
\sin(t)\longmapsto \cos(t)-s\cdot\sin(t)
$$
which is equivalent to $f(x)$ satisfying
$$
x\longmapsto \sqrt{1-x^2}-s\cdot x
$$
on the other hand, considering the second coordinate function, $f$ satisfies
$$
x\longmapsto \sqrt{1-x^2}+s\cdot x
$$
Substitute $s$ above by a continuous (possible differentiable) function of $x$. Then this is only consistent if $s(\sin(t))=-s(\cos(t))$. In particular we must have, since $\sin(0.25\pi)=\cos(0.25\pi)=0.5\sqrt 2$ that $s(0.5\sqrt 2)=0$. Then just pick your favourite function $s(x)$ with these properties. Differentiable, if you like.
A: The answer is no, this equation has infinitely many solutions. For example
$$
f_{\lambda}(x)=\left\{\matrix{\sqrt{1-x^2}+\lambda\frac{x}{\sqrt{1-x^2}}\left(1-\frac{x}{\sqrt{1-x^2}}\right)&\hbox{ if}&0\leq x\leq \frac{1}{\sqrt{2}},\cr
\cr
\sqrt{1-x^2}+\lambda\left(\frac{\sqrt{1-x^2}}{x}-1\right)\qquad\quad\quad&\hbox{ if}&\frac{1}{\sqrt{2}}\leq x\leq 1.
}\right.
$$
is a solution for every $\lambda$. 
Indeed, note that the quantity
$$G(t)=\cos t\, f_\lambda (\sin t)+\sin t\, f_\lambda(\cos t)$$ satisfy $G(t)=G(\frac{\pi}{2}-t)$, so, it is enough to show that $G(t)=1$ for $0\leq t\leq\frac{\pi}{4}$. For these values of $t$ we have $0\leq \sin t\leq\frac{1}{\sqrt 2}\leq\cos t\leq 1$, hence
$$
\eqalign{
f_\lambda(\sin t)&=\cos t+\lambda \tan t\,\left(1-\tan t\right)\cr
f_\lambda(\cos t)&=\sin t+\lambda \left(\tan t-1\right)
}
$$
So
$$
G(t)=1+\lambda(\sin t\,(1-\tan t)+\sin t\,(\tan t-1))=1.
$$
and the verification is complete.
$\qquad\square$
