ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2563 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากสูตรมุมสามเท่า} \\ \sin 3\theta = 3\sin \theta -4{\sin }^3\theta \\ \cos 3\theta = 4{\cos }^2\theta -3\cos \theta \\ \mathrm{จากโจทย์} \\ \frac{{\sin }^{2}3\theta }{{\sin }^2\theta }-\frac{{\cos }^{2}3\theta }{{\cos }^2\theta } = 1 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{{\left(3\sin \theta -4{\sin }^2\theta \right)}^2}{{\sin }^2\theta }-\frac{{\left(4{\cos }^3\theta -3\cos \theta \right)}^2}{{\cos }^2\theta } = 1 \\ \frac{{\left(\sin \theta \left(3-4{\sin }^2\theta \right)\right)}^2}{{\sin }^2\theta }-\frac{{\left(\cos \theta \left(4{\cos }^2\theta -3\right)\right)}^2}{{\cos }^2\theta } = 1 \end{gathered}$$

$$\begin{gathered} \frac{{\sin }^2\theta {\left(3-4{\sin }^2\theta \right)}^2}{{\sin }^2\theta }-\frac{{\cos }^2\theta {\left(4{\cos }^2\theta -3\right)}^2}{{\cos }^2\theta } = 1 \\ {\left(3-4{\sin }^2\theta \right)}^2-{\left(4{\cos }^2\theta -3\right)}^2 = 1 \\ \left[\left(3-4{\sin }^2\theta \right)-\left(4{\cos }^2\theta -3\right)\right]\left[\left(3-4{\sin }^2\theta \right)+\left(4{\cos }^2\theta -3\right)\right] = 1 \end{gathered}$$

$$\begin{gathered} \left(6-4{\sin }^2\theta -4{\sin }^2\theta \right)\left(-4{\sin }^2\theta +4{\cos }^2\theta \right) = 1 \\ \left(6-4\left({\sin }^2\theta +{\cos }^2\theta \right)\right)\left(4\right)\left({\cos }^2\theta -{\sin }^2\theta \right) = 1 \\ \left(6-4\left(1\right)\right)\left(4\right)\left({\cos }^2\theta -\left(-\left(1-{\cos }^2\theta \right)\right)\right) = 1 \end{gathered}$$

$$\begin{gathered} -1+2{\cos }^2\theta = \frac{1}{8} \\ {\cos }^2\theta = \frac{9}{16} \\ \cos \theta = \frac{3}{4} \end{gathered}$$

$$\begin{gathered} \mathrm{เนื่องจาก} \theta \in \left(0, \frac{\mathrm{\pi }}{2}\right) \cos \mathrm{ต้องเป็นบวก} \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 5 \end{gathered}$$