ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{สูตรตรีโกณ} \\ \cos A+\cos B = 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right) \\ \sin A+\sin B = 2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right) \\ \mathrm{จากโจทย์} \frac{{\left(a+b\right)}^2}{a^2+b^2} \mathrm{ตรงกับข้อใด} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{{\left(a+b\right)}^2}{a^2+b^2} = \frac{a^2+2ab+b^2}{a^2+b^2} \\ = \frac{\left(a^2+b^2\right)+2ab}{a^2+b^2} \\ = \frac{\cancel{a^2+b^2}}{\cancel{a^2+b^2}}+\frac{2ab}{a^2+b^2} \end{gathered}$$

$$\begin{gathered} = 1 +\frac{2ab}{a^2+b^2} \\ = 1+\frac{2ab}{{\displaystyle \frac{ab}{ab}\left(a^2+b^2\right)}} \\ = 1+\frac{2\cancel{ab}}{{\displaystyle \cancel{ab}\left(\frac{a}{b}+\frac{b}{a}\right)}} \\ = 1+\frac{2}{{\displaystyle \frac{a}{b}+\frac{b}{a}}} \to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} a = \cos 15+\cos 50 \\ a = \cos 50+\cos 15 \\ a = 2\cos \left(\frac{50+15}{2}\right)\cos \left(\frac{50-15}{2}\right) \\ a = 2\cos \left(\frac{65}{2}\right)\cos \left(\frac{35}{2}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} b = \sin 15+\sin 50 \\ b = \sin 50+\sin 15 \\ b = 2\sin \left(\frac{50+15}{2}\right)\cos \left(\frac{50-15}{2}\right) \\ b = 2\sin \left(\frac{65}{2}\right)\cos \left(\frac{35}{2}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{หาค่า} \frac{a}{b} = \frac{2\cos \left({\displaystyle \frac{65}{2}}\right)\cancel{\cos \left({\displaystyle \frac{35}{2}}\right)}}{2\sin \left(\frac{65}{2}\right)\cancel{\cos \left(\frac{35}{2}\right)}} \\ \frac{a}{b} = \frac{2\cos \left({\displaystyle \frac{65}{2}}\right)}{2\sin \left(\frac{65}{2}\right)} ; \mathrm{แทนใน} \boxed{1} \\ \mathrm{หาค่า} \frac{b}{a} = \frac{2\sin \left({\displaystyle \frac{65}{2}}\right)\cancel{\cos \left({\displaystyle \frac{35}{2}}\right)}}{2\cos \left(\frac{65}{2}\right)\cancel{\cos \left(\frac{35}{2}\right)}} \\ \frac{b}{a} = \frac{2\sin \left({\displaystyle \frac{65}{2}}\right)}{2\cos \left(\frac{65}{2}\right)} ; \mathrm{แทนใน} \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{จะได้} \\ \frac{{\left(a+b\right)}^2}{a^2+b^2} = 1+\frac{2}{{\displaystyle \frac{a}{b}}+{\displaystyle \frac{b}{a}}} \\ = 1+\frac{2}{{\displaystyle \frac{\cos \left({\displaystyle \frac{65}{2}}\right)}{\sin \left(\frac{65}{2}\right)}}+\frac{\sin \left({\displaystyle \frac{65}{2}}\right)}{\cos \left(\frac{65}{2}\right)}} \\ = 1+\frac{2\sin \left(\frac{65}{2}\right)\cos \left(\frac{65}{2}\right)}{{\displaystyle {\cos }^2\left(\frac{{\displaystyle 65}}{{\displaystyle 2}}\right)+{\sin }^2\left(\frac{65}{2}\right)}} \end{gathered}$$

$$\begin{gathered} \mathrm{โดย} {\sin }^2\theta +{\cos }^2\theta = 1 \\ = 1+\frac{2\sin \left(\frac{65}{2}\right)\cos \left(\frac{65}{2}\right)}{{\displaystyle 1}} \\ \mathrm{โดย} 2\sin \theta \cos \theta =\sin 2\theta \\ = 1+\sin 65° \\ = 1+\cos \left(90°-65°\right) \\ = 1+\cos 25° \end{gathered}$$