ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2560 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{สูตรตรีโกณ} \sin A+\sin B = 2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right) \\ \left|u\times v\right| = \left|u\right|\left|v\right|\sin \theta \\ u·v = \left|u\right|\left|v\right|\cos \theta \to \boxed{a} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \overline{v}=\left(sin80°+sin20°\right)\overline{i}+a\overline{j}+b\overline{k} \\ =\left[2sin\left(\frac{80+20}{2}\right)cos\left(\frac{80-20}{2}\right)\right]\overline{i}+a\overline{j}+b\overline{k} \\ =\left(2sin50°cos30°\right)\overline{i}+a\overline{j}+b\overline{k} \\ =\left(2sin50°\frac{\sqrt{3}}{2}\right)i+aj+bk \\ =\left(\sqrt{3}sin50°\right)i+aj+bk \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \left|v\times i\right|=sin70°+sin10° \\ \left|v\right|\left|i\right|sin\theta =2sin\left(\frac{70+10}{2}\right)cos\left(\frac{70-10}{2}\right) ; \mathrm{โดย} \left|i\right|=1 \\ \left|v\right|sin\theta =2sin40°cos30° \\ \left|v\right|sin\theta =2sin40°\frac{\sqrt{3}}{2} \\ \left|v\right|sin\theta =\sqrt{3}sin40° \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{ยกกำลัง} 2 \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ \hspace{0.17em}\hspace{0.17em} {\left|v\right|}^{2}si{n}^2\theta =3si{n}^{2}40°\to \boxed{1} \\ \mathrm{ต้องหา} {\left|v\right|}^2{\cos }^2\theta \mathrm{เพื่อเข้าสูตร} {\sin }^2\theta +{\cos }^2\theta =1 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{a} v·i=\left|v\right|\left|i\right|cos\theta ; \mathrm{โดย} \left|i\right|=1 \\ \mathrm{จะได้} \left[\begin{array}{c}\sqrt{3}sin50°\\ a\\ b\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]=\left|v\right|\left(1\right)cos\theta \\ \sqrt{3}sin50°=\left|v\right|cos\theta ; \mathrm{ยกกำลัง} 2 \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ {\left|v\right|}^{2}co{s}^2\theta =3si{n}^{2}50°\to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{นำ} \boxed{1}+\boxed{2} {\left|v\right|}^{2}si{n}^2\theta +{\left|v\right|}^{2}co{s}^2\theta =3si{n}^{2}40°+3si{n}^{2}50° \\ \mathbf{โดย} sin\theta =cos\left(90°-\theta \right) ; \\ {\left|v\right|}^2\left[si{n}^2\theta +co{s}^2\theta \right]=3si{n}^{2}40°+3co{s}^{2}40° \\ \mathbf{โดย} si{n}^2\theta +co{s}^2\theta =1 ; \\ {\left|v\right|}^2\left(1\right)=3\left(si{n}^{2}40°+co{s}^{2}40°\right) \\ \mathbf{โดย} si{n}^{2}40°+co{s}^{2}40°=1 ; \\ {\left|v\right|}^2=3\left(1\right) \\ \mathrm{ดังนั้น} {\left|v\right|}^2=3 \end{gathered}$$