Question

Find the maximum and minimum values of the trigonometric expression $12\sin \theta -5\cos \theta$

Answer 1

Hint: Divide and multiply the expression given in the question by 13 and take $\cos \alpha ={\displaystyle \frac{12}{13}}$ . Also, use the fact that the range of the sine function is [-1,1] and is defined for all real numbers.

Complete step-by-step answer:
Now we will start with the simplification of the expression that is given in the question.
$12\sin \theta -5\cos \theta$
Now we divide and multiply the expression by 13. On doing so, we get
$13({\displaystyle \frac{12}{13}}\sin \theta -{\displaystyle \frac{5}{13}}\cos \theta )$
We take ${\displaystyle \frac{12}{13}}$ to be equal to $\cos \alpha$, then we $\sin \alpha$ can be calculated as:
$\begin{array}{rl}& {\sin }^2\alpha +{\cos }^2\alpha =1\\ & \Rightarrow \sin \alpha =\sqrt{1-{\cos }^2\alpha }=\sqrt{1-{\displaystyle \frac{144}{169}}}={\displaystyle \frac{5}{13}}\end{array}$
Therefore, using the above assumption and result in the expression $13({\displaystyle \frac{12}{13}}\sin \theta -{\displaystyle \frac{5}{13}}\cos \theta )$ , we get
$13({\displaystyle \frac{12}{13}}\sin \theta -{\displaystyle \frac{5}{13}}\cos \theta )$
$=13(cos\alpha sin\theta -\sin \alpha \cos \theta )$
Now using the formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$ , our expression becomes:
$=13\sin (\theta -\alpha )$
Now we let $$\theta -\alpha =\beta$$ . Therefore, we get our final expression to be $13\sin \beta$ .
We know that the sine function can have a maximum value of 1 and minimum value of -1. Also, our final expression is maximum when $\sin \beta$ is maximum and minimum when $\sin \beta$ is minimum. The expression is minimum.
Therefore, the maximum and minimum value of the expression $12\sin \theta -5\cos \theta$ is 13 and -13, respectively.


Note: If you want, you can directly remember that the maximum and minimum value of the expression $a\sin \theta -b\cos \theta$ is $\sqrt{a^2+b^2}\text{ and }-\sqrt{a^2+b^2}$ , respectively. Also, for solving the above question, you can use the method of derivative, but that would be difficult to solve and would require a good hold on the concepts of inverse trigonometric functions.