ข้อสอบ PAT 1 - มีนาคม 2558

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em}\arctan \left(\frac{2\cos 10°-\cos 50°}{\sin 70°-\cos 80°}\right) \\ \mathbf{หาค่า}\mathbf{ }\frac{2\cos 10°-\cos 50°}{\sin 70°-\cos 80°} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\cos 10°+\left(\cos 10°-\cos 50°\right)}{\sin 70°-\cos 80°}\to \boxed{1} \\ \mathrm{จากสูตร} \mathrm{cosA}-\mathrm{cosB}=-2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \end{gathered}$$
$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{จะได้} =\frac{\cos \left(90°-80°\right)+\left(-2\sin {\left({\displaystyle \frac{10-50}{2}}\right)}^{°}\sin {\left(\frac{10+50}{2}\right)}^{°}\right)}{\sin 70°-\cos \left(90-10\right)°} \\ =\frac{\sin 80°-2\sin \left(-20\right)°\sin 30°}{\sin 70°-\sin 10°} \\ =\frac{\sin 80°+2\sin 20°\left({\displaystyle \frac{1}{2}}\right)}{\sin 70°-\sin 70}\to \boxed{2} \\ \mathrm{จากสูตร} \mathrm{sinA}+\mathrm{sinB}=2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \\ \mathrm{และ} \mathrm{sinA}-\mathrm{sinB}=2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} \mathrm{จะได้} =\frac{2\sin {\left({\displaystyle \frac{80+20}{2}}\right)}^{°}\cos {\left({\displaystyle \frac{80-20}{2}}\right)}^{°}}{2\cos {\left({\displaystyle \frac{70+10}{2}}\right)}^{°}\sin {\left({\displaystyle \frac{70-10}{2}}\right)}^{°}} \\ =\frac{2\sin 50°\cos 30°}{2\cos 40°\sin 30°} \\ =\frac{\sin 50°}{\cos \left(90-50\right)°}\cot 30° \\ =\frac{\sin 50°}{\sin 50°}\cot \left(90-60\right)° \\ =\tan 60° \\ =\sqrt{3} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \arctan \left(\frac{2\cos 10°-\cos 50°}{\sin 70°-\cos 80°}\right)=\arctan \left(\sqrt{3}\right) \\ =\frac{\mathrm{\pi }}{3}=60° \end{gathered}$$