ข้อสอบ PAT 1 - ตุลาคม 2559

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em}13\sin \left(2\arctan \frac{2}{3}\right)+4{\tan }^2\left(\arccos \frac{2}{3}\right)\to \boxed{1} \\ \mathrm{กำหนดให้} \mathrm{A}=\arctan \frac{2}{3} \mathrm{B}=\arccos \frac{2}{3} \\ \mathrm{tanA}=\frac{2}{3} \mathrm{cosB}=\frac{2}{3} \\ -\mathrm{นำมาวาดรูปสามเหลี่ยม} \end{gathered}$$


$$\begin{gathered} \mathrm{จะได้} \mathrm{sinA}=\frac{2}{\sqrt{13}} \mathrm{tanB}=\frac{\sqrt{5}}{2} \\ \mathrm{cosA}=\frac{3}{\sqrt{3}} \\ \mathrm{จาก} \boxed{1} \mathrm{ค่าของ} 13\sin \left(2\arctan \frac{2}{3}\right)+4{\tan }^2\left(\arccos \frac{2}{3}\right) ; \mathrm{แทน} \mathrm{A},\mathrm{B} \\ \mathrm{จะได้} =13\sin \left(2\mathrm{A}\right)+4{\tan }^2\mathrm{B} \\ =13\left(2\mathrm{sinAcosA}\right)+4{\tan }^2\mathrm{B} \\ =13\left(2\left(\frac{2}{\sqrt{13}}\right)\left(\frac{3}{\sqrt{13}}\right)\right)+4{\left(\frac{\sqrt{5}}{2}\right)}^2 \end{gathered}$$

$$\begin{gathered} =\cancel{13}\left(\frac{12}{\cancel{13}}\right)+\cancel{4}\left(\frac{5}{\cancel{4}}\right) \\ =12+5 \\ =17 \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ค่าของ} 13\sin \left(2\arctan \frac{2}{3}\right)+4{\tan }^2\left(\arccos \frac{2}{3}\right)=17 \end{gathered}$$