ข้อสอบ PAT 1 - เมษายน 2557

$$\begin{gathered} \underline{\mathrm{พิจารณาข้อ} \left(\mathrm{ก}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em} \cos 2\mathrm{B}=2{\cos }^2\left(45°+\mathrm{A}\right) \\ =2{\left(\cos 45\cos \mathrm{A}-\sin 45 \sin \mathrm{A}\right)}^2 \\ =2(\frac{\sqrt{2}}{2}\cos \mathrm{A}-\frac{\sqrt{2}}{2}\sin \mathrm{A}{)}^2 \\ =2{\left(\frac{\sqrt{2}}{2}\right)}^2{\left(\cos \mathrm{A}-\sin \mathrm{A}\right)}^2 \\ =2\left(\frac{2}{4}\right)\left({\cos }^2\mathrm{A}-2\sin \mathrm{A} \cos \mathrm{A}+{\sin }^2\mathrm{A}\right) \\ =1-2\sin \mathrm{A} \cos \mathrm{A} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} {\sin }^2\mathrm{B}=\sin \mathrm{A} \cos \mathrm{A} \\ \mathrm{จะได้} \cos 2\mathrm{B}=1-2{\sin }^2\mathrm{B} \\ =\cos 2\mathrm{B} \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$​

$$\begin{gathered} \underline{\mathrm{พิจารณา} }\underline{\mathrm{ข้อ} \left(\mathrm{ข}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em} \sin \mathrm{A}=\sqrt{2}\sin \mathrm{B}\to \boxed{\mathrm{a}} ; \mathrm{ยกกำลัง} 2 \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ \mathrm{จะได้} {\sin }^2\mathrm{A}=2{\sin }^2\mathrm{B}\to \boxed{1} \\ \mathrm{จากโจทย์}\hspace{0.17em} \sqrt{3}\sec \mathrm{B}=\sqrt{2}\sec \mathrm{A} \\ \frac{\sqrt{3}}{\cos \mathrm{B}}=\frac{\sqrt{2}}{\cos \mathrm{A}} \\ \sqrt{3}\cos \mathrm{A}=\sqrt{2}\cos \mathrm{B} ; \mathrm{ยกกำลัง} 2 \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ 3{\cos }^2\mathrm{A}=2{\cos }^2\mathrm{B}\to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{นำ} \boxed{1}+\boxed{2} {\sin }^2\mathrm{A}+3{\cos }^2\mathrm{A}=2{\sin }^2\mathrm{B}+2{\cos }^2\mathrm{B} \\ \left(1-{\cos }^2\mathrm{A}\right)+3{\cos }^2\mathrm{A}=2\left({\sin }^2\mathrm{B}+{\cos }^2\mathrm{B}\right) \\ 1+2{\cos }^2\mathrm{A}=2\left(1\right) \\ {\cos }^2\mathrm{A}=\frac{1}{2} \\ \cos \mathrm{A}=\frac{1}{\sqrt{2}} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} 0\le \mathrm{A},\mathrm{B}\le \frac{\mathrm{\pi }}{2} \\ \mathrm{จะได้} \mathrm{A}=45° \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{\mathrm{a}} \sin \mathrm{A}=\sqrt{2}\sin \mathrm{B} \\ \mathrm{จะได้} \sin 45°=\sqrt{2}\sin \mathrm{B} \\ \frac{\sin 45°}{\sqrt{2}}=\sin \mathrm{B} \\ \frac{1}{2}=\sin \mathrm{B} \\ \mathrm{จะได้} \mathrm{B}=30° \\ \mathrm{จากโจทย์} \sin 10\mathrm{A}+\cos 10\mathrm{B}=0.5 \\ \mathrm{แทนค่า} \mathrm{A} ,\mathrm{B} \end{gathered}$$
$$\begin{gathered} \mathrm{จะได้} \sin 450°+\cos 300°=0.5 \\ \sin \left(360°+90°\right)+\cos \left(360°-60°\right)=0.5 \\ \sin 90°+\cos 60°=0.5 \\ 1+\frac{1}{2}=0.5 \\ 1.5\ne 1.5 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ผิด} \end{gathered}$$