ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากสูตร} \sin 2\mathrm{A}=2\sin \mathrm{A} \cos \mathrm{A}\hspace{0.17em} \\ \cos 2\mathrm{A}={\cos }^2\mathrm{A}-{\sin }^2\mathrm{A} \\ \sin \left(\mathrm{A}-\mathrm{B}\right)=\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B} \\ \mathrm{จากโจทย์}\hspace{0.17em}\left(\mathrm{ก}\right) \frac{\cos 10°- \sin 10°}{\cos 10°+\sin 10°} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{\cos 10°- \sin 10°}{\cos 10°+\sin 10°}·\left(\frac{\cos 10°- \sin 10°}{\cos 10°- \sin 10°}\right) \\ =\frac{{\cos }^{2}10°-2\sin 10° \cos 10°+{\sin }^{2}10°}{{\cos }^{2}10°- {\sin }^{2}10°} \\ =\frac{1-\sin 20°}{\cos 20°}=\frac{1}{\cos 20°}-\frac{\sin 20°}{\cos 20°} \\ = \sec 20° - \tan 20° \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \left(\mathrm{ข}\right) \sqrt{3}\cot 20° \\ \mathrm{จะได้} =\frac{\sqrt{3}\cos 20°}{\sin 20°} \\ =\frac{2\left({\displaystyle \frac{\sqrt{3}}{2}}\cos 20°\right)}{\sin 20°} \\ =\frac{2\left({\displaystyle \frac{\sqrt{3}}{2}\cos 20°-\frac{1}{2}\sin 20°+\frac{1}{2}\sin 20°}\right)}{\sin 20°} \\ =\frac{2\left({\displaystyle \left(\sin 60° \cos 20°-\cos 60° \sin 20°\right)+\frac{1}{2}\sin 20°}\right)}{\sin 20°} \end{gathered}$$
$$\begin{gathered} =\frac{2\left({\displaystyle \sin 40°+\frac{1}{2}\sin 20°}\right)}{\sin 20°} \\ =\frac{2\left({\displaystyle 2\sin 20° \cos 20°+\frac{1}{2}\sin 20°}\right)}{\sin 20°} \\ =\frac{4{\displaystyle \sin }{\displaystyle 20}{\displaystyle °}{\displaystyle }{\displaystyle \cos }{\displaystyle 20}{\displaystyle °}{\displaystyle +}{\displaystyle \sin }{\displaystyle 20}{\displaystyle °}}{\sin 20°} \\ =\frac{{\displaystyle \cancel{\sin 20°} \left(4\cos 20°+1\right)}}{\cancel{\sin 20°}} \\ = 1+4\cos 20° \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ถูก} \end{gathered}$$