ข้อสอบคณิต 9 สามัญ - ปี 2555

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{cosA}-\sqrt{2}\mathrm{cosB}=0 \\ \cos A=\sqrt{2}\cos B\to \boxed{1} \\ \mathrm{จากโจทย์}\hspace{0.17em} \cos 2\mathrm{A}+3\cos 2\mathrm{B}=-2 \\ \mathrm{จะได้} \left(2{\cos }^2\mathrm{A}-1\right)+3\left(2{\cos }^2\mathrm{B}-1\right)=-2 \\ \mathrm{จาก} \boxed{1} \mathrm{cosA}=\sqrt{2}\mathrm{cosB} \mathrm{จะได้} \end{gathered}$$

$$\begin{gathered} \left[2{\left(\sqrt{2}\mathrm{cosB}\right)}^2-1\right]+3\left(2{\cos }^2\mathrm{B}-1\right)=-2 \\ \left(4{\cos }^2\mathrm{B}-1\right)+6{\cos }^2\mathrm{B}-3=-2 \\ 10{\cos }^2\mathrm{B}-4=-2 \\ {\cos }^2\mathrm{B}=\frac{2}{10} \\ {\cos }^2\mathrm{B}=\frac{1}{5} \\ \mathrm{cosB}=\frac{1}{\sqrt{5}} \\ \mathrm{แสดงว่า}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{sinB}=\frac{2}{\sqrt{5}} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \mathrm{cosA}=\sqrt{2}\mathrm{cosB} \\ \mathrm{cosA}=\sqrt{2}·\frac{1}{\sqrt{5}} \\ \mathrm{cosA}=\frac{\sqrt{2}}{\sqrt{5}}\to \mathrm{sinA}=\frac{\sqrt{3}}{\sqrt{5}} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{สามเหลี่ยม} \mathrm{ABC} \mathrm{มีมุม} \mathrm{A} \mathrm{และมุม} \mathrm{B} \mathrm{เป็นมุมแหลม} \\ \mathrm{จะได้} \mathrm{A}+\mathrm{B}+\mathrm{C}=180° \\ \mathrm{C}=180-\left(\mathrm{A}+\mathrm{B}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \cos \mathrm{C}=\cos \left[180-\left(\mathrm{A}+\mathrm{B}\right)\right] \\ =-\cos \left(\mathrm{A}+\mathrm{B}\right) \\ =-\left[\mathrm{cosAcosB}-\mathrm{sinAsinB}\right] \\ =-\mathrm{cosAcosB}+\mathrm{sinAsinB} \\ =-\frac{\sqrt{2}}{\sqrt{5}}·\frac{1}{\sqrt{5}}+\frac{\sqrt{3}}{\sqrt{5}}·\frac{2}{\sqrt{5}} \\ =-\frac{\sqrt{2}}{5}+\frac{2\sqrt{3}}{5} \\ =\frac{1}{5}\left(2\sqrt{3}-\sqrt{2}\right) \end{gathered}$$