Question

Find the maximum and minimum values of each of the following trigonometrical expressions:
(i) $12sin\theta -5cos\theta$
(ii) $12cos\theta +5sin\theta +4$
(iii) $5cos\theta +3sin(\frac{\pi }{6}-\theta )+4$
(iv) $sin\theta -cos\theta +1$

Answer 1

Let $f\left(\theta \right)=12sin\theta -5cos\theta$
We know that
$-\sqrt{(12{)}^2+(-5{)}^2}\le f\left(\theta \right)\le \sqrt{(12{)}^2+(-5{)}^2}$
$\Rightarrow -\sqrt{144+25}\le f\left(\theta \right)\le \sqrt{144+25}\Rightarrow -\sqrt{169}\le f\left(\theta \right)\le \sqrt{169}\Rightarrow -13\le f\left(\theta \right)\le 13$
Hence, minimum and maximum values of $12sin\theta -5cos\theta$ are -13 and 13 respectively.
(ii) Let $f\left(\theta \right)=12cos\theta +5cos\theta +4$
We know that
$-\sqrt{(12{)}^2+(5{)}^2}\le 12cos\theta +5sin\theta \le \sqrt{(12{)}^2+(-5{)}^2}$
$\Rightarrow -\sqrt{144+25}\le 12cos\theta +5sin\theta \le \sqrt{144+25}\Rightarrow -\sqrt{169}\le 12cos\theta +5sin\theta \le \sqrt{169}$
$\Rightarrow -13\le 12cos\theta +5sin\theta \le 13\Rightarrow -13+4\le 13+4$
$\Rightarrow -9\le 12cos\theta +5sin\theta +4\le 17\Rightarrow -9\le f\left(0\right)\le 17$
Hence, minimum and maximum values of 12 cos \theta+5 sin \theta +4 are -9 and 17 respectively.
(iii) Let $f\left(\theta \right)=5cos\theta +3sin(\frac{\pi }{6}-\theta )+4$
Then, $f\left(\theta \right)=5cos\theta +3(sin\frac{\pi }{6}-cos\frac{\pi }{6}sin\theta )+4$
$=5cos\theta +3[\frac{1}{2}cos\theta -\frac{\sqrt{3}}{2}sin\theta ]+4-5cos\theta +\frac{3}{2}cos\theta -\frac{\sqrt{3}}{2}sin\theta +4$
$=(5+\frac{3}{2})cos\theta -\frac{3\sqrt{3}}{2}sin\theta +4=\frac{13}{2}cos\theta -\frac{3\sqrt{3}}{2}sin\theta +4=\frac{13}{2}cos\theta -\left(\frac{-3\sqrt{3}}{2}\right)sin\theta +4$
We know that,
$-\sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{-3\sqrt{3}}{2}\right)}^2}\le \frac{13}{2}cos\theta -\left(\frac{-3\sqrt{3}}{2}\right)sin\theta \le \sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{-3\sqrt{3}}{2}\right)}^2}$
$\Rightarrow -\sqrt{\frac{169}{4}+\frac{27}{4}}\le \frac{13}{2}cos\theta -(\frac{-3\sqrt{3}}{2}sin\theta \le \sqrt{\frac{169}{4}}+\frac{27}{4})$
$\Rightarrow -\sqrt{\frac{196}{4}}\le \frac{13}{2}cos\theta -\left(\frac{-3\sqrt{3}}{2}sin\theta \right)\le \sqrt{\frac{196}{4}}\Rightarrow -\frac{14}{2}\le \frac{13}{2}cos\theta -\frac{3\sqrt{3}}{2}sin\theta \le \frac{14}{2}$
$\Rightarrow -7\le \frac{13}{2}cos\theta -\frac{3\sqrt{3}}{2}sin\theta \le 7$
$\Rightarrow -7+4\le \frac{13}{2}cos\theta -\frac{3\sqrt{3}}{2}sin\theta +4\le 7+4$
$\Rightarrow -3\le \frac{13}{2}cos\theta -\left(\frac{-3\sqrt{3}}{2}\right)sin\theta +4\le 11\Rightarrow -3\le f\left(\theta \right)\le 11$
Hence, -3 and 11 are respectively the minimum and the maximum values of $5cos\theta +3sin(\frac{\pi }{4}-\theta )+4$. (iv) Let $f\left(\theta \right)=sin\theta -cos\theta +1$ Then,
$f\left(\theta \right)=sin\theta +(-1)cos\theta +1$
$=(-1)cos\theta +sin\theta +1$
We know that
$-\sqrt{(-1{)}^2+(1{)}^2}\le -cos\theta +sin\theta \le \sqrt{(-1{)}^2}+(1{)}^2$
$\Rightarrow -\sqrt{1+1}\le -cos\theta +sin\theta \le \sqrt{1+1}$
$\Rightarrow -\sqrt{2}\le -cos\theta +sin\theta \le \sqrt{2}$
$\Rightarrow -\sqrt{2}+1\le -cos\theta +sin\theta +1\le \sqrt{2}+1\Rightarrow 1-\sqrt{2}\le f\left(\theta \right)\le 1+\sqrt{2}$
Hence, minimum and maximum values of $sin\theta -cos\theta +1$ are $1-\sqrt{2}$ and $1+\sqrt{2}$ respectively.