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 $\cos \alpha =\dfrac{12}{13}$ . 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
$13\left( \dfrac{12}{13}\sin \theta -\dfrac{5}{13}\cos \theta \right)$
We take $\dfrac{12}{13}$ to be equal to $\cos \alpha$, then we $\sin \alpha$ can be calculated as:
$\begin{aligned} & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1 \\\ & \Rightarrow \sin \alpha =\sqrt{1-{{\cos }^{2}}\alpha }=\sqrt{1-\dfrac{144}{169}}=\dfrac{5}{13} \\\ \end{aligned}$
Therefore, using the above assumption and result in the expression $13\left( \dfrac{12}{13}\sin \theta -\dfrac{5}{13}\cos \theta \right)$ , we get
$13\left( \dfrac{12}{13}\sin \theta -\dfrac{5}{13}\cos \theta \right)$
$=13\left( cos\alpha sin\theta -\sin \alpha \cos \theta \right)$
Now using the formula $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ , our expression becomes:
$=13\sin \left( \theta -\alpha \right)$
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.