Question

Reduce each of the following expressions to the sine and cosine of a single expression:

(i) $\sqrt{3}\sin \mathrm{\theta }-\cos \mathrm{\theta }$
(ii) cos θ − sin θ
(iii) 24 cos θ + 7 sin θ

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}f\left(\theta \right)=\sqrt{3}\sin \theta -\cos \theta \\ \mathrm{Dividing}\mathrm{and}\mathrm{multiplying}\mathrm{by}\sqrt{3+1},\mathrm{i}.\mathrm{e}.\mathrm{by}2,\mathrm{we}\mathrm{get}: \\ f\left(\theta \right)=2\left(\frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta \right) \\ \Rightarrow f\left(\theta \right)=2\left(\cos \frac{\mathrm{\pi }}{6}\sin \theta -\sin \frac{\mathrm{\pi }}{6}\cos \theta \right) \\ \Rightarrow f\left(\theta \right)=2\sin \left(\theta -\frac{\mathrm{\pi }}{6}\right) \\ \mathrm{Again}, \\ f\left(\theta \right)=2\left(\frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta \right) \\ \Rightarrow f\left(\theta \right)=2\left(\sin \frac{\mathrm{\pi }}{3}\sin \theta -\cos \frac{\mathrm{\pi }}{3}\cos \theta \right) \\ \Rightarrow f\left(\theta \right)=-2\cos \left(\frac{\mathrm{\pi }}{3}+\theta \right) \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}f\left(\theta \right)=\cos \theta -\sin \theta \\ \mathrm{Dividing}\mathrm{and}\mathrm{multiplying}\mathrm{by}\sqrt{1^2+1^2},\mathrm{i}.\mathrm{e}.\mathrm{by}\sqrt{2,}\mathrm{we}\mathrm{get}: \\ f\left(\theta \right)=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\sin \theta \right) \\ \Rightarrow f\left(\theta \right)=\sqrt{2}(\cos 45°\cos \theta -\sin 45°\sin \theta ) \\ \Rightarrow f\left(\theta \right)=\sqrt{2}\cos \left(\frac{\mathrm{\pi }}{4}+\theta \right) \\ \mathrm{Again}, \\ f\left(\theta \right)=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\sin \theta \right) \\ \Rightarrow f\left(\theta \right)=\sqrt{2}(\sin 45°\cos \theta -\cos 45°\sin \theta ) \\ \Rightarrow f\left(\theta \right)=\sqrt{2}\sin \left(\frac{\mathrm{\pi }}{4}-\theta \right) \end{gathered}$$


$$\begin{gathered} \left(iii\right)\mathrm{Let}f\left(\theta \right)=24\cos \theta +7\sin \theta \\ \mathrm{Dividing}\mathrm{and}\mathrm{multiplying}\mathrm{by}\sqrt{{24}^2+7^2},\mathrm{i}.\mathrm{e}.\mathrm{by}25,\mathrm{we}\mathrm{get}: \\ f\left(\theta \right)=25\left(\frac{24}{25}\cos \theta +\frac{7}{25}\sin \theta \right) \\ \Rightarrow f\left(\theta \right)=25(\sin \alpha \cos \theta +\cos \alpha \sin \theta ),\mathrm{where}\sin \alpha =\frac{24}{25}\mathrm{and}\cos \alpha =\frac{7}{25} \\ \Rightarrow f\left(\theta \right)=25\sin (\alpha +\theta ),\mathrm{where}\tan \alpha =\frac{24}{7}. \\ \mathrm{Again}, \\ f\left(\theta \right)=25\left(\frac{24}{25}\cos \theta +\frac{7}{25}\sin \theta \right) \\ \Rightarrow f\left(\theta \right)=25(\cos \alpha \cos \theta +\sin \alpha \sin \theta ),\mathrm{where}\cos \alpha =\frac{24}{25},\sin \alpha =\frac{7}{25}. \\ \Rightarrow f\left(\theta \right)=25\cos (\alpha -\theta ),\mathrm{where}\tan \alpha =\frac{7}{24}. \end{gathered}$$