ข้อสอบ PAT 1 - ตุลาคม 2555

$$\begin{gathered} \underline{\mathrm{ตัวเลือกที่} 1} \\ \mathrm{สูตรตรีโกณ} \tan \left(\mathrm{A}-\mathrm{B}\right)=\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A} \tan \mathrm{B}}\to \boxed{1} \\ \mathrm{จากโจทย์} \cos 75°=\left(2-\sqrt{3}\right)\cos 15° \\ \sin 15°=\left(2-\sqrt{3}\right)\cos 15° \\ \frac{\sin 15°}{\cos 15°}=2-\sqrt{3} \\ \tan 15°=2-\sqrt{3} \\ \tan \left(60°-45°\right)=2-\sqrt{3} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{จะได้} \frac{\tan 60°-\tan 45°}{1+\tan 60° \tan 45°}=2-\sqrt{3} \\ \frac{\sqrt{3}-1}{ 1+\sqrt{3}\left(1\right)}=2-\sqrt{3} \\ \frac{\sqrt{3}-1}{ \sqrt{3}+1}=2-\sqrt{3} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{คูณ} \sqrt{3}-1 \mathrm{ทั้งเศษและส่วน} \\ \frac{\sqrt{3}-1}{ \sqrt{3}+1}·\left(\frac{\sqrt{3}-1}{ \sqrt{3}-1}\right)=2-\sqrt{3} \end{gathered}$$
$$\begin{gathered} \frac{3-2\sqrt{3}+1}{ 3-1}=2-\sqrt{3} \\ \frac{4-2\sqrt{3}}{ 2}=2-\sqrt{3} \\ \frac{2\left(2-\sqrt{3}\right)}{ 2}=2-\sqrt{3} \\ 2-\sqrt{3}=2-\sqrt{3} \mathrm{ตัวเลือกที่} 1 \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{ตัวเลือกที่} 2} \\ \mathrm{สูตรตรีโกณ} \sin \mathrm{A}+\sin \mathrm{B}=2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\to \boxed{2} \\ \mathrm{จากโจทย์} \cos 10°+\sin 40° = \cos 20° \\ \mathbf{โดย} \cos \mathrm{A}=\sin \left(90°-\mathrm{A}\right) ; \\ \sin 80°+\sin 40°=\cos 20° \\ \mathrm{จาก} \boxed{2} \mathrm{จะได้} 2\sin \left(\frac{80+40}{2}\right)\cos \left(\frac{80-40}{2}\right)=\cos 20° \\ 2\sin 60°\cos 20°=\cos 20° \\ 2\left(\frac{\sqrt{3}}{2}\right)\cos 20°=\cos 20° \\ \sqrt{3}\cos 20°\ne \cos 20° \\ \mathrm{ตัวเลือกที่} 2 \mathrm{ผิด} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{ตัวเลือกที่} 3} \\ \mathrm{สูตรตรีโกณ} \cos \mathrm{A}+\cos \mathrm{B}=2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\to \boxed{3} \\ \cos \mathrm{A}-\cos \mathrm{B}=-2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\to \boxed{4} \\ \mathrm{จากโจทย์}\hspace{0.17em} \frac{\tan 3\mathrm{A}}{\mathrm{cotA}}=\frac{\cos {\displaystyle }{\displaystyle 2}{\displaystyle \mathrm{A}}{\displaystyle +}{\displaystyle \cos }{\displaystyle }{\displaystyle 4}{\displaystyle \mathrm{A}}}{\cos {\displaystyle }{\displaystyle 2}{\displaystyle \mathrm{A}}{\displaystyle -}{\displaystyle \cos }{\displaystyle }{\displaystyle 4}{\displaystyle \mathrm{A}}} \\ \mathrm{จาก} \boxed{3} , \boxed{4} =\frac{2\cos \left({\displaystyle \frac{2\mathrm{A}+4\mathrm{A}}{2}}\right)\cos \left({\displaystyle \frac{2\mathrm{A}-4\mathrm{A}}{2}}\right)}{-2\sin \left({\displaystyle \frac{2\mathrm{A}+4\mathrm{A}}{2}}\right)\sin \left({\displaystyle \frac{2\mathrm{A}-4\mathrm{A}}{2}}\right)} \end{gathered}$$

$$\begin{gathered} =\frac{2\cos 3\mathrm{A} \cos \left(-\mathrm{A}\right)}{-2\sin 3\mathrm{A} \sin \left(-\mathrm{A}\right)} \\ =\frac{\cos 3\mathrm{A} \mathrm{cosA}}{-\sin 3\mathrm{A} \left(-\mathrm{sinA}\right)} \\ =\left(\frac{\cos 3\mathrm{A}}{\sin 3\mathrm{A}}\right)\left(\frac{\mathrm{cosA}}{\mathrm{sinA}}\right) \\ =\cot 3\mathrm{A} \cot \mathrm{A} \\ =\frac{\cot 3\mathrm{A} }{\mathrm{tanA}} \mathrm{ตัวเลือกที่} 3 \mathrm{ผิด} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{ตัวเลือกที่} 4} \\ \mathrm{สูตรตีโกณ} 2\sin {\mathrm{\theta }}_1 \cos {\mathrm{\theta }}_2=\sin \left({\mathrm{\theta }}_1+{\mathrm{\theta }}_2\right)+\sin \left({\mathrm{\theta }}_1-{\mathrm{\theta }}_2\right)\to \boxed{5} \\ \mathrm{จากโจทย์} \sin 2\mathrm{A}+\cos 2\mathrm{B}=2 \sin \left(\mathrm{A}-\mathrm{B}\right) \cos \left(\mathrm{A}+\mathrm{B}\right) \\ \mathrm{จาก} \boxed{5} \mathrm{จะได้} \\ =\sin \left[\left(\mathrm{A}-\mathrm{B}\right)+\left(\mathrm{A}+\mathrm{B}\right)\right]+\sin \left[\left(\mathrm{A}-\mathrm{B}\right)-\left(\mathrm{A}+\mathrm{B}\right)\right] \\ =\sin 2\mathrm{A}+\sin \left(-2\mathrm{B}\right) ; \sin \left(-\mathrm{\theta }\right)=-\sin \mathrm{\theta } \\ =\sin 2\mathrm{A}-\sin 2\mathrm{B} \mathrm{ตัวเลือกที่} 4 \mathrm{ผิด} \end{gathered}$$