Olympiad Maths Prep

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Problem 981

AIME late
Algebra Difficulty 5.9 Find the answer

8.401. {tan⁡x2+tan⁡y2=2cot⁡x+cot⁡y=−1.8.\left\{\begin{array}{l}\tan \frac{x}{2}+\tan \frac{y}{2}=2 \\ \cot x+\cot y=-1.8 .\end{array}\right.

Official solution

## Solution.

Domain of definition: {sin⁡x≠0,sin⁡y≠0.\left\{\begin{array}{l}\sin x \neq 0, \\ \sin y \neq 0 .\end{array}\right.

Using the formula ctg⁡α=1−tg⁡2α22tg⁡α2\operatorname{ctg} \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{2 \operatorname{tg} \frac{\alpha}{2}}, we can write the system of equations as

{tg⁡y2=2−tg⁡x21−tg⁡2x22tg⁡x2+1−tg⁡2y22tg⁡y2=−1.8⇒tg⁡y2=2−tg⁡x2,1−tg⁡2x22tg⁡x2+1−(2−tg⁡2x2)22(2−tg⁡x2)=−1.8;4tg⁡2x2−8tg⁡x2−5=0. \begin{aligned} & \left\{\begin{array}{l} \operatorname{tg} \frac{y}{2}=2-\operatorname{tg} \frac{x}{2} \\ \frac{1-\operatorname{tg}^{2} \frac{x}{2}}{2 \operatorname{tg} \frac{x}{2}}+\frac{1-\operatorname{tg}^{2} \frac{y}{2}}{2 \operatorname{tg} \frac{y}{2}}=-1.8 \end{array} \Rightarrow \operatorname{tg} \frac{y}{2}=2-\operatorname{tg} \frac{x}{2},\right. \\ & \frac{1-\operatorname{tg}^{2} \frac{x}{2}}{2 \operatorname{tg} \frac{x}{2}}+\frac{1-\left(2-\operatorname{tg}^{2} \frac{x}{2}\right)^{2}}{2\left(2-\operatorname{tg} \frac{x}{2}\right)}=-1.8 ; 4 \operatorname{tg}^{2} \frac{x}{2}-8 \operatorname{tg} \frac{x}{2}-5=0 . \end{aligned}

Solving this equation as a quadratic equation in tg⁡x2\operatorname{tg} \frac{x}{2}, we get tg⁡x2=−12,tg⁡x2=52⋅\operatorname{tg} \frac{x}{2}=-\frac{1}{2}, \operatorname{tg} \frac{x}{2}=\frac{5}{2} \cdot Further, we have

1) {tg⁡y12=52,tg⁡x12=−12 or 2) {tg⁡y22=−12tg⁡x22=52\left\{\begin{array}{l}\operatorname{tg} \frac{y_{1}}{2}=\frac{5}{2}, \\ \operatorname{tg} \frac{x_{1}}{2}=-\frac{1}{2} \text { or 2) }\end{array}\left\{\begin{array}{l}\operatorname{tg} \frac{y_{2}}{2}=-\frac{1}{2} \\ \operatorname{tg} \frac{x_{2}}{2}=\frac{5}{2}\end{array}\right.\right.
2) {x12=−arctg⁡12+πk1,y12=arctg⁡52+πk2,⇔{x1=−2arctg⁡12+2πk1,y1=2arctg⁡52+2πk2.\left\{\begin{array}{l}\frac{x_{1}}{2}=-\operatorname{arctg} \frac{1}{2}+\pi k_{1}, \\ \frac{y_{1}}{2}=\operatorname{arctg} \frac{5}{2}+\pi k_{2},\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{1}, \\ y_{1}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{2} .\end{array}\right.\right.
3) {x22=arctg⁡52+πk1,y22=−arctg⁡12+πk2,⇔{x2=2arctg⁡52+2πk1,y2=−2arctg⁡12+2πk2,\left\{\begin{array}{l}\frac{x_{2}}{2}=\operatorname{arctg} \frac{5}{2}+\pi k_{1}, \\ \frac{y_{2}}{2}=-\operatorname{arctg} \frac{1}{2}+\pi k_{2},\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{2}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{1}, \\ y_{2}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{2},\end{array}\right.\right. where k1k_{1} and k2∈Zk_{2} \in Z.

Answer: x1=−2arctg⁡12+2πk1,y1=2arctg⁡52+2πk2x_{1}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{1}, y_{1}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{2};

x2=2arctg⁡52+2πk1,y2=−2arctg⁡12+2πk2, where k1 and k2∈Z x_{2}=2 \operatorname{arctg} \frac{5}{2}+2 \pi k_{1}, y_{2}=-2 \operatorname{arctg} \frac{1}{2}+2 \pi k_{2}, \text { where } k_{1} \text { and } k_{2} \in Z

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.