Simplify the following expression:

$$
\sqrt{2-\cos \vartheta-\sin \vartheta+\sqrt{3(1-\cos \vartheta)(1+\cos \vartheta-2 \sin \vartheta)}} .
$$
I. solution. Notice that if we multiply the first factor in parentheses by 3 under the inner square root, the sum of the two factors is twice the expression before the inner square root, and thus

$$
\begin{gathered}
\frac{3}{2}(1-\cos \vartheta)+\frac{1}{2}(1+\cos \vartheta-2 \sin \vartheta)=2-\cos \vartheta-\sin \vartheta \\
K^{2}=\left(\sqrt{3 \cdot \frac{1-\cos \vartheta}{2}}+\sqrt{\frac{1+\cos \vartheta-2 \sin \vartheta}{2}}\right)^{2}
\end{gathered}
$$

Further transforming the expressions under the square roots,

$$
\frac{1}{2}(1+\cos \vartheta-2 \sin \vartheta)=\cos ^{2} \frac{\vartheta}{2}-2 \sin \frac{\vartheta}{2} \cos \frac{\vartheta}{2}=\cos ^{2} \frac{\vartheta}{2}\left(1-2 \operatorname{tg} \frac{\vartheta}{2}\right)
$$

therefore,

$$
K=\sqrt{3} \cdot\left|\sin \frac{\vartheta}{2}\right|+\left|\cos \frac{\vartheta}{2}\right| \cdot \sqrt{1-2 \operatorname{tg} \frac{\vartheta}{2}}
$$

The expression before the inner square root is positive:

$$
(1-\cos \vartheta)+(1-\sin \vartheta)>0
$$

since no negative number can stand in either parenthesis, therefore (1) and (3) are not defined where the expression under the inner square root is

$$
1+\cos \vartheta-2 \sin \vartheta\frac{1}{2}
$$

This is true for the tangent function in one period, for example in the $(0, \pi)$ interval, if

$$
\operatorname{arctg} \frac{1}{2}<\frac{\vartheta}{2}<\frac{\pi}{2}
$$

or in the $(0,2 \pi)$ interval for the following values:

$$
2 \operatorname{arctg} \frac{1}{2}=\operatorname{arctg} \frac{4}{3}<\vartheta<\pi
$$

However, it is more convenient to give the domain of definition in the $(-\pi, \pi)$ interval, because in this interval it is coherent: for $-\pi \leqq \vartheta \leqq \operatorname{arc} \operatorname{tg} 4 / 3$, (1) can be written in the form (3), adding that for $\vartheta=-\pi$, the second term of (3) is 0 due to the first form of (2).

Imre Bárány (Budapest, Corvinus High School) László Surányi (Budapest, Fazekas M. High School)

Remark. In fact, we used the identity

$$
\sqrt{A+\sqrt{B}}=\sqrt{\left(A+\sqrt{A^{2}-B}\right) / 2}+\sqrt{\left(A-\sqrt{A^{2}-B}\right) / 2}
$$

In our case,

$$
\begin{aligned}
A^{2}-B & =(2-\cos \vartheta-\sin \vartheta)^{2}-3(1-\cos \vartheta)(1+\cos \vartheta-2 \sin \vartheta)= \\
& =(1-2 \cos \vartheta+\sin \vartheta)^{2}
\end{aligned}
$$

II. solution. Let's try introducing the trigonometric functions of $\vartheta / 2=\alpha$, so that the expressions $1-\cos \vartheta, 1+\cos \vartheta$ can be transformed into one square term each:

$$
\begin{aligned}
K & =\sqrt{2-\cos \vartheta-\sin \vartheta+\sqrt{3\left[1-\left(1-2 \sin ^{2} \alpha\right)\right] \cdot\left[1+\left(2 \cos ^{2} \alpha-1\right)-4 \sin \alpha \cos \alpha\right]}}= \\
& =\sqrt{2-\cos \vartheta-\sin \vartheta+2 \sqrt{3 \sin ^{2} \alpha\left(\cos ^{2} \alpha-2 \sin \alpha \cos \alpha\right)}}= \\
& =\sqrt{2-\cos ^{2} \alpha+\sin ^{2} \alpha-2 \sin \alpha \cos \alpha+2 \sqrt{3} \cdot|\sin \alpha| \cdot \sqrt{\cos ^{2} \alpha-2 \sin \alpha \cos \alpha}}= \\
& =\sqrt{3 \sin ^{2} \alpha+\left(\cos ^{2} \alpha-2 \sin \alpha \cos \alpha\right)+2 \sqrt{3} \cdot|\sin \alpha| \cdot \sqrt{\cos ^{2} \alpha-2 \sin \alpha \cos \alpha}}= \\
& =\sqrt{\left(\sqrt{3} \cdot|\sin \alpha|+\sqrt{\cos ^{2} \alpha-2 \sin \alpha \cos \alpha}\right)^{2}}= \\
& =\sqrt{3} \cdot|\sin \alpha|+\sqrt{\cos ^{2} \alpha-2 \sin \alpha \cos \alpha}=\sqrt{3} \cdot|\sin \alpha|+|\cos \alpha| \cdot \sqrt{1-2 \operatorname{tg} \alpha}
\end{aligned}
$$

which is identical to (3).

Remark. We recommend comparing this problem with problem 1306. 2[^0]

[^0]:    ${ }^{1}$ See for example László Faragó: Collection of Mathematical Problems, 3rd edition, Tankönyvkiadó, Budapest 1965. p. 13, problem 80.

    ${ }^{2}$ See the solution in K. M. L. 29 (1964/12 issue) p. 209.