ข้อสอบคณิต 9 สามัญ - ปี 2562

$$\begin{gathered} \mathrm{สูตร} a^2-b^2 = \left(a-b\right)\left(a+b\right) \\ a^4-b^4 = \left(a^2-b^2\right)\left(a^2+b^2\right) \\ \mathrm{สูตรตรีโกณ} \cos 2\theta = {\cos }^2\theta -{\sin }^2\theta \\ {\sin }^2\theta +{\cos }^2\theta = 1 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} {\cos }^4\left(\frac{5\mathrm{\pi }}{12}\right)-{\sin }^4\left(\frac{5\mathrm{\pi }}{12}\right) \\ \mathrm{กำหนดให้} \theta = \frac{5\mathrm{\pi }}{12} \\ \mathrm{จะได้} {\cos }^4\left(\frac{5\mathrm{\pi }}{12}\right)-{\sin }^4\left(\frac{5\mathrm{\pi }}{12}\right) \end{gathered}$$

$$\begin{gathered} = {\cos }^4\theta -{\sin }^4\theta \\ = \left({\cos }^2\theta -{\sin }^2\theta \right)\left({\cos }^2\theta +{\sin }^2\theta \right) \\ = \left(\cos 2\theta \right)\left(1\right) ; \mathrm{แทน} \theta = \frac{5\mathrm{\pi }}{12} \end{gathered}$$

$$\begin{gathered} = \cos \left(2·\frac{5\mathrm{\pi }}{12}\right) \\ = \cos \frac{5\mathrm{\pi }}{6} = \cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right) \\ = -\cos \frac{\mathrm{\pi }}{6} = -\frac{\sqrt{3}}{2} \end{gathered}$$