ข้อสอบ PAT 1 - มีนาคม 2560

$\mathrm{สูตรตรีโกณ} {\sin }^2\theta +{\cos }^2\theta =1 \to {\sin }^2\theta =1-{\cos }^2\theta$

$$\begin{gathered} \mathrm{จากโจทย์} 2{\sin }^2\theta = 3\cos \theta \\ 2\left(1-{\cos }^2\theta \right) = 3\cos \theta \\ 2-2{\cos }^2\theta = 3\cos \theta \\ 2{\cos }^2\theta +3\cos \theta -2 = 0 \\ \left(2\cos \theta -1\right)\left(\cos \theta +2\right) = 0 \\ \mathrm{จะได้} 2\cos \theta -1 = 0 \mathrm{หรือ} \cos \theta +2=0 \\ \cos \theta = \frac{1}{2} \cos \theta =-2 \\ \theta = 60° \mathrm{เป็นไปไม่ได้} \mathrm{เพราะ} -1\le \cos \le 1 \\ \mathrm{ดังนั้น} \theta = 60° \mathrm{เท่านั้น} \end{gathered}$$

$$\begin{gathered} \mathrm{ค่าของ} \cos e{c}^2\left(\frac{\mathrm{\pi }}{2}-\theta \right){\cos }^2\theta +\frac{\tan \theta }{\cos ec2\theta } \\ -\mathrm{แทน} \theta =60° \\ \mathrm{จะได้} = \cos e{c}^2\left(90°-60°\right){\cos }^{2}60°+\frac{\tan 60°}{\cos ec\left[2\left(60°\right)\right]} \\ = \cos e{c}^2\left(30°\right){\cos }^{2}60°+\frac{\tan 60°}{\cos ec120°} \end{gathered}$$
$$\begin{gathered} \mathbf{โดย} \cos ec\theta = \frac{1}{\sin \theta } \\ = \cos e{c}^{2}30°{\cos }^{2}60°+\tan 60°\left(\sin 120°\right) \\ \mathbf{โดย} \sin 120° = \sin 60° \\ = \frac{{\cos }^{2}60°}{{\sin }^{2}30°}+\tan 60°\left(\sin 60°\right) \end{gathered}$$

$$\begin{gathered} = \frac{{\left({\displaystyle \frac{1}{2}}\right)}^2}{{\left(\frac{1}{2}\right)}^2}+\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) \\ = 1+\frac{3}{2} =1+1.5 \\ = 2.5 \\ \mathrm{ดังนั้น} \cos e{c}^2\left(\frac{\mathrm{\pi }}{2}-\theta \right){\cos }^2\theta +\frac{\tan \theta }{\cos ec2\theta } = 2.5 \end{gathered}$$