Let's determine the maximum of the function

$$
f(t)=A \sin t + B \sin 2t
$$

considering only the non-negative values of \( f(t) \). ( \( A \) and \( B \) are constants.)
Let's assume that $A$ and $B$ are not 0, because if either is 0, we would already be dealing with a well-known function's maximum.

Express $f(t)$ using the trigonometric functions of $t$:

$$
\mathrm{f}(t)=A \sin t+B \sin 2 t=A \sin t+2 B \sin t \cos t=\sin t \cdot(A+2 B \cos t)
$$

Since we are only examining the values of $f(t) \geq 0$, we can instead examine the function $g(t)=f^{2}(t)$. $f(t)$ and $g(t)$ achieve their maximum at the same points.

$$
\begin{aligned}
& \mathrm{g}(t)=\sin ^{2} t(A+2 B \cos t)^{2}=\left(1-\cos ^{2} t\right)(A+2 B \cos t)^{2}= \\
= & (1-\cos t)(1+\cos t)(A+2 B \cos t)(A+2 B \cos t)
\end{aligned}
$$

The location of the maximum of the function does not change if we multiply the first and second factors by a positive $m$ and $n$ respectively. Let's try to choose these in such a way that the sum of the four factors is independent of $t$, i.e., constant. In this case, $g(t)$ achieves its maximum when the factors are equal.

The sum is

$$
(4 B-m+n) \cos t+m+n+2 A
$$

This is independent of $t$ if

$$
4 B-m+n=0
$$

The factors are equal, for example, if

$$
m(1-\cos t)=A+2 B \cos t
$$

from which

$$
m=\frac{A+2 B \cos t}{1-\cos t}
$$

and

$$
n(1+\cos t)=A+2 B \cos t
$$

from which

$$
n=\frac{A+2 B \cos t}{1+\cos t}
$$

Substitute the values of $n$ and $m$ into (1):

$$
\frac{A+2 B \cos t}{1+\cos t}-\frac{A+2 B \cos t}{1-\cos t}+4 B=0
$$

Rearranging:

$$
4 B \cos ^{2} t+A \cos t-2 B=0
$$

The roots of the quadratic equation are

$$
(\cos t)_{1,2}=\frac{-A \pm \sqrt{A^{2}+32 B^{2}}}{8 B}
$$

Thus, $g(t)$ - and therefore $f(t)$ - achieves its maximum at the above values of $\cos t$.

At that point, the maximum is:

$$
\begin{aligned}
& \mathrm{f}(t)_{\max }=A \sin t(A+2 B \cos t)= \pm \sqrt{1-\cos ^{2} t}(A+2 B \cos t)= \\
& = \pm \sqrt{1-\frac{\left(-A \pm \sqrt{A^{2}+32 B^{2}}\right)^{2}}{64 B^{2}}}\left(A+2 B \cdot \frac{-A \pm \sqrt{A^{2}+32 B^{2}}}{8 B}\right)= \\
& = \pm \frac{\sqrt{32 B^{2} \pm 2 A \sqrt{A^{2}+32 B^{2}}-2 A^{2}} \cdot\left(3 A \pm \sqrt{A^{2}+32 B^{2}}\right)}{32 B}= \\
& = \pm \frac{\sqrt{\left(3 A \pm \sqrt{A^{2}+32 B^{2}}\right)\left(-A \pm \sqrt{A^{2}+32 B^{2}}\right)}\left(3 A \pm \sqrt{A^{2}+32 B^{2}}\right)}{32 B}= \\
& = \pm \frac{\sqrt{\left(3 A \pm \sqrt{A^{2}+32 B^{2}}\right)^{3}\left(-A \pm \sqrt{A^{2}+32 B^{2}}\right)}}{32 B}
\end{aligned}
$$

If $B$ is positive, then the positive sign before the fraction, and if $B$ is negative, then the negative sign is valid. The value under the square root must be positive (or possibly 0). Therefore, the sign under the square root must be chosen based on the values of $A$ and $B$ so that the signs of the two factors are always the same.

Balint Pödör (Bp. II., Rakoczi g. III. o. t.)