Question

Find the maximum and minimum values of each of the following trigonometrical expressions:

(i) 12 sin θ − 5 cos θ
(ii) 12 cos θ + 5 sin θ + 4
(iii) $5\cos \mathrm{\theta }+3\sin \left(\frac{\mathrm{\pi }}{6}-\mathrm{\theta }\right)+4$
(iv) sin θ − cos θ + 1

Answer 1

(i)
$$\begin{gathered} \mathrm{Le}tf\left(\theta \right)=12\sin \theta -5\cos \theta \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ -\sqrt{{12}^2+(-5{)}^2}\le 12\sin \theta -5\cos \theta \le \sqrt{{12}^2+(-5{)}^2} \\ -\sqrt{144+25}\le 12\sin \theta -5\cos \theta \le \sqrt{144+25} \\ -13\le 12\sin \theta -5\cos \theta \le 13 \\ \mathrm{Hence}\mathrm{the}\mathrm{maximum}\mathrm{and}\mathrm{minumun}\mathrm{values}\mathrm{of}f\left(\theta \right)\mathrm{are}13\mathrm{and}-13,\mathrm{respectively}. \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Let}f\left(\theta \right)=12\cos \theta +5\sin \theta +4 \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ -\sqrt{{12}^2+5^2}\le 12\cos \theta +5\sin \theta \le \sqrt{{12}^2+5^2}\mathrm{for}\mathrm{all}\theta \\ \Rightarrow -\sqrt{169}\le 12\cos \theta +5\sin \theta \le \sqrt{169} \\ \Rightarrow -13\le 12\cos \theta +5\sin \theta \le 13 \\ \Rightarrow -9\le 12\cos \theta +5\sin \theta +4\le 17 \\ \mathrm{Hence},\mathrm{the}\mathrm{maximum}\mathrm{and}\mathrm{minimum}\mathrm{vaues}\mathrm{of}f\left(\theta \right)are17\mathrm{and}-9,\mathrm{respectively}. \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{Let}f\left(\theta \right)=5\cos \theta +3\sin \left(\frac{\mathrm{\pi }}{6}-\theta \right)+4 \\ \mathrm{Now}f\left(\theta \right)=5\cos \theta +3\left(\sin 30°\cos \theta -\cos 30°\sin \theta \right)+4 \\ =5\cos \theta +\frac{3}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta +4 \\ =\frac{13}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta +4 \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ -\sqrt{{\left(\frac{13}{2}\right)}^2+{\left(-\frac{3\sqrt{3}}{2}\right)}^2}\le \frac{13}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta \le \sqrt{{\left(\frac{13}{2}\right)}^2+{\left(-\frac{3\sqrt{3}}{2}\right)}^2}forall\theta \\ \mathrm{Therefore}, \\ -\sqrt{\frac{169+27}{4}}\le \frac{13}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta \le \sqrt{\frac{169+27}{4}} \\ \Rightarrow -\frac{14}{2}+4\le \frac{13}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta +4\le \frac{14}{2}+4 \\ \Rightarrow -3\le \frac{13}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta +4\le 11 \\ \mathrm{Hence},\mathrm{maximum}\mathrm{and}\mathrm{minimun}\mathrm{values}\mathrm{of}f\left(\theta \right)are11\mathrm{and}-3,\mathrm{respectively}. \end{gathered}$$

(iv)
$$\begin{gathered} \mathrm{Let}f\left(\theta \right)=\sin \theta -\cos \theta +1 \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ -\sqrt{1^2+(-1{)}^2}\le \sin \theta -\cos \theta \le \sqrt{1^2+(-1{)}^2}\mathrm{for}\mathrm{all}\theta \\ \Rightarrow -\sqrt{2}\le \sin \theta -\cos \theta \le \sqrt{2} \\ \Rightarrow -\sqrt{2}+1\le \sin \theta -\cos \theta +1\le \sqrt{2}+1 \\ \mathrm{Hence}\mathrm{maximum}\mathrm{and}\mathrm{minimum}\mathrm{values}\mathrm{of}f\left(\theta \right)are1+\sqrt{2}\mathrm{and}1-\sqrt{2},\mathrm{respectively}. \end{gathered}$$