ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2566 (หลักสูตรใหม่)

$$\begin{gathered} \mathrm{จากโจทย์} \tan \left(arc\cos \left(\frac{5}{13}\right)+arc\sin \left(\frac{3}{5}\right)\right) \\ \mathrm{กำหนดให้} A = arc\cos \left(\frac{5}{13}\right) \\ \mathrm{จะได้} \cos A = \frac{5}{13} \\ - \mathrm{วาดรูปสามเหลี่ยม} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \tan A = \frac{12}{5} \to \boxed{1} \\ \mathrm{กำหนดให้} B = arc\sin \left(\frac{3}{5}\right) \\ \mathrm{จะได้} \sin B = \frac{3}{5} \\ - \mathrm{วาดรูปสามเหลี่ยม} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \tan B = \frac{3}{4} \to \boxed{2} \\ \mathrm{จากสูตร} \tan \left(A+B\right) = \frac{\tan A+\tan B}{1-\tan A·\tan B} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \tan \left(arc\cos \left(\frac{5}{13}\right)+arc\sin \left(\frac{3}{5}\right)\right) \mathrm{เท่ากับเท่าใด} \\ \mathrm{จะได้} \tan \left(arc\cos \left(\frac{5}{13}\right)+arc\sin \left(\frac{3}{5}\right)\right) = \tan \left(A+B\right) \end{gathered}$$

$$\begin{gathered} = \frac{\tan A+\tan B}{1-\tan A·\tan B} ; \mathrm{นำ} \boxed{1} , \boxed{2} \mathrm{แทนค่า} \\ = \frac{{\displaystyle \frac{12}{5}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{12}{5}}·{\displaystyle \frac{3}{4}}} = \frac{{\displaystyle \frac{48+15}{20}}}{{\displaystyle \frac{20-36}{20}}} \end{gathered}$$

$$\begin{gathered} = \frac{{\displaystyle \frac{63}{20}}}{{\displaystyle \frac{-16}{20}}} = -\frac{63}{16} \\ \mathrm{ดังนั้น} \tan (arc\cos \left(\frac{5}{13}\right)+arc\sin \left(\frac{3}{5}\right))= -\frac{63}{16} \end{gathered}$$