ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathbf{สูตร} \sin \left(\mathrm{A}+\mathrm{B}\right)=\sin \mathrm{A} \cos \mathrm{B} +\cos \mathrm{A} \sin \mathrm{B} \\ 2\sin \mathrm{A} \sin \mathrm{B}=\cos \left(\mathrm{A}-\mathrm{B}\right)-\cos \left(\mathrm{A}+\mathrm{B}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} 2{\sin }^{2}60°(\tan 5°+\tan 85°) - 12\sin 70° \\ =2{\left(\frac{\sqrt{3}}{2}\right)}^2\left(\frac{\sin 5°}{\cos 5°}+\frac{\sin 85°}{\cos 85°}\right)-12\sin 70° \\ =2\left(\frac{3}{4}\right)\left(\frac{\sin 5°\cos 85°+\sin 85°\cos 5°}{\cos 5°\cos 85°}\right)-12\sin 70° \\ =\frac{3}{2}\left(\frac{\sin \left(5°+ 85°\right)}{\cos 5°\sin 5°}\right)-12\sin 70° \end{gathered}$$

$$\begin{gathered} =\frac{3\sin 90°}{2\sin 5°\cos 5°}-12\sin 70° ; \sin 90°=1 \\ =\frac{3}{\sin 10°}-12\sin 70° \\ =\frac{3-12\sin 70°\sin 10°}{\sin 10°} \end{gathered}$$

$$\begin{gathered} =\frac{3-6\left(2\sin 70°\sin 10°\right)}{\sin 10°} \\ =\frac{3-6\left(\cos \left(70°-10°\right)-\cos \left(70°+10°\right)\right)}{\sin 10°} \\ =\frac{3-6\left(\cos 60°-\cos 80°\right)}{\sin 10°} \end{gathered}$$

$$\begin{gathered} =\frac{3-6\cos 60°+6\cos 80°}{\sin 10°} \\ =\frac{3-6\left({\displaystyle \frac{1}{2}}\right)+6\sin 10°}{\sin 10°} \\ =\frac{6\cancel{\sin 10°}}{\cancel{\sin 10°}} \\ =6 \end{gathered}$$