ข้อสอบ PAT 1 - มีนาคม 2557

$$\begin{gathered} \mathrm{จากโจทย์} cot\left(arc\cos \sqrt{\frac{2}{3}}-arc\cos \frac{1+\sqrt{6}}{2\sqrt{3}}\right) \\ \mathrm{ให้} A=arc\cos \sqrt{\frac{2}{3}} \mathrm{และ} B=arc\cos \frac{1+\sqrt{6}}{2\sqrt{3}} \\ \cos A=\sqrt{\frac{2}{3}} \cos B=\frac{1+\sqrt{6}}{2\sqrt{3}} \end{gathered}$$


    $$\begin{gathered} a=\sqrt{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} b=\sqrt{{\left(2\sqrt{3}\right)}^2-{\left(1+\sqrt{6}\right)}^2} \\ =1 =\sqrt{12-\left(1+2\sqrt{6}+6\right)} \\ =\sqrt{5-2\sqrt{6}}\to \sqrt{3}-\sqrt{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \tan A=\frac{1}{\sqrt{2}} \tan B=\frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}} \\ \mathrm{จากโจทย์} cot\left(arc\cos \sqrt{\frac{2}{3}}-arc\cos \frac{1+\sqrt{6}}{2\sqrt{3}}\right) \\ -\mathrm{แทนค่า} A , B \\ \mathrm{จะได้} cot\left(A-B\right) \\ \mathrm{จากสูตร} cot\left(A-B\right)=\frac{1}{\tan \left(A-B\right)} \\ \mathrm{โดย} \tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A·\tan B} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} cot\left(A-B\right)=\frac{1+\tan A·\tan B}{\tan A-tan B} \\ -\mathrm{แทนค่า} \tan A , \tan B \\ \mathrm{จะได้} cot \left(A-B\right)=\frac{1+({\displaystyle \frac{1}{\sqrt{2}}}·{\displaystyle \frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}})}{{\displaystyle \frac{1}{\sqrt{2}}}-\left({\displaystyle \frac{\sqrt{3}-\sqrt{2}}{1+\sqrt{6}}}\right)} \\ =\frac{\sqrt{2}\left(1+\sqrt{6}\right)+\sqrt{3}-\sqrt{2}}{\left(1+\sqrt{6}\right)-\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)} \end{gathered}$$

$$\begin{gathered} =\frac{\cancel{\sqrt{2}}+\sqrt{12}+\sqrt{3}-\cancel{\sqrt{2}}}{1+\cancel{\sqrt{6}}-\cancel{\sqrt{6}}+2} \\ =\frac{\sqrt{12}+\sqrt{3}}{3}=\frac{2\sqrt{3}+\sqrt{3}}{3} \\ =\sqrt{3} \\ \mathrm{ดังนั้น} cot\left(arc\cos \sqrt{\frac{2}{3}}-arc\cos \frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\sqrt{3} \end{gathered}$$