ข้อสอบ PAT 1 - มีนาคม 2559

$$\begin{gathered} \mathrm{จากโจทย์} A = arc\sin \left(\frac{\sin \theta }{\sqrt{1+{\sin }^2\theta }}\right) \\ \mathrm{จะได้} sin A = \frac{sin \theta }{\sqrt{1+si{n}^2\theta }} \end{gathered}$$

$$\begin{gathered} \mathrm{จากรูป} tan A = sin \theta \\ \mathrm{จากโจทย์} B = arc\tan \left(1-\sin \theta \right) \\ \mathrm{จะได้} tan B = 1-sin \theta \\ \mathrm{จากโจทย์} C = arc\tan \sqrt{\sin \theta -{\sin }^2 \theta } \\ \mathrm{จะได้} tan C = \sqrt{sin \theta -si{n}^2 \theta } \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} A+B = 2C \\ \mathrm{take} \tan \mathrm{ทั้งสองข้างสมการ} \mathrm{จะได้} \\ \mathrm{จะได้} \tan \left(A+B\right) = \tan \left(2C\right) \\ \frac{\tan A+\tan B}{1-\tan A \tan B} = \frac{2\tan C}{1-{\tan }^2 C} \\ \frac{\left[\sin \theta \right]+\left[1-\sin \theta \right]}{1-(\sin \theta )\left(1-\sin \theta \right)} = \frac{2\sqrt{\sin \theta -{\sin }^2 \theta }}{1-\left(\sin \theta -{\sin }^2 \theta \right)} \\ \frac{1}{\cancel{1-\sin \theta +{\sin }^2 \theta }} = \frac{2\sqrt{\sin \theta -{\sin }^2 \theta }}{\cancel{1-\sin \theta +{\sin }^2 \theta }} \end{gathered}$$

$$\begin{gathered} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ส่วนเท่ากันทั้ง} 2 \mathrm{ข้าง} \mathrm{ตัดทิ้งได้} \\ 1 = 2\sqrt{\sin \theta -{\sin }^2 \theta } \\ 1 = 4\left(\sin \theta -{\sin }^2 \theta \right) \\ 4{\sin }^2 \theta -4\sin \theta +1 = 0 \\ {\left(2\sin \theta -1\right)}^2 = 0 \\ \sin \theta = \frac{1}{2} ; \mathrm{โดย} 0<\theta <90° \\ \theta = 30° \end{gathered}$$​

$$\begin{gathered} \mathrm{ดังนั้น} 3 {\sin }^4\theta +{\cos }^4\theta = 3{\sin }^{4}30°+{\cos }^{4}30° \\ = 3{\left(\frac{1}{2}\right)}^4+{\left(\frac{\sqrt{3}}{2}\right)}^4 \\ = \frac{12}{16} = \frac{3}{4} \\ = 0.75 \\ \mathbf{ค่าของ}\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{ }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{4}}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{4}}\mathit{\theta }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{75} \end{gathered}$$