ข้อสอบ PAT 1 - มีนาคม 2559

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

$$\begin{gathered} \mathrm{จากโจทย์} 4\sin 40°-\tan 40° \\ \mathrm{จะได้} 4\sin 40°-\tan 40° \\ = 4\sin 40°-\frac{\sin 40°}{\cos 40°} \\ = \frac{ 4\sin 40°\cos 40°-\sin 40°}{\cos 40°} \\ = \frac{2\left(2\sin 40°\cos 40°\right)-\sin 40°}{\cos 40°} \\ = \frac{2\sin 80°-\sin 40°}{\cos 40°} \\ = \frac{\sin 80°+\sin 80°-\sin 40°}{\cos 40°} \end{gathered}$$

$$\begin{gathered} = \frac{\sin 80°+2\cos \left({\displaystyle \frac{80+40}{2}}\right)\sin \left({\displaystyle \frac{80-40}{2}}\right)}{\cos 40°} \\ = \frac{\sin 80°+2\cos 60°\sin 20°}{\cos 40°} \\ = \frac{\sin 80°+2\left({\displaystyle \frac{1}{2}}\right)\sin 20°}{\cos 40°} \end{gathered}$$

$$\begin{gathered} = \frac{\sin 80°+\sin 20°}{\cos 40°} \\ = \frac{2\sin \left({\displaystyle \frac{80+20}{2}}\right)\cos \left(\frac{80-20}{2}\right)}{\cos 40°} \\ = \frac{2\sin 50°\cos 30°}{\cos 40°} \\ \mathrm{โดยที่} \sin 50° = \sin \left(90-40\right) = \cos 40° \\ = \frac{2\cos 40°\cos 30°}{\cos 40°} \\ = 2\cos 30° = 2\left(\frac{\sqrt{3}}{2}\right) \\ = \sqrt{3} \end{gathered}$$

$$\begin{gathered} \mathbf{พิจารณาตัวเลือก} \\ \mathrm{ตัวเลือก} 1) \cos 405° = \cos \left(360+45\right) \\ = \cos 45° = \frac{\sqrt{2}}{2} \\ \mathrm{ตัวเลือก} 2) \sin 420° = \sin \left(360+60\right) \\ = \sin 60° = \frac{\sqrt{3}}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{ตัวเลือก} 3) sec \left(-60°\right) = sec 60° \\ = \frac{1}{\cos 60°} = 2 \\ \mathrm{ตัวเลือก} 4) \tan \left(-120°\right) = -\tan 120° \\ = -\left(-\tan 60°\right) = \sqrt{3} \\ \mathrm{ตัวเลือก} 5) cot\left(-135°\right) = -cot135° \\ = -\left(-cot45°\right) = 1 \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 4) \tan \left(-120°\right) \end{gathered}$$​