# Given that 7 cos x + 24 sin x = R cos (x - theta) where R > 0 and theta is acute, how do you find the minimum and maximum values of 7 cos x + 24 sin x - 12 and the corresponding values of x?

Aug 4, 2018

$\min . \left\{7 \cos x + 24 \sin x - 12\right\} = - 37 , \mathmr{and} ,$

$\max . \left\{7 \cos x + 24 \sin x - 12\right\} = 13$.

#### Explanation:

Given that, $7 \cos x + 24 \sin x = R \cos \left(x - \theta\right) , R > 0 , \theta \text{ acute}$.

We expanding $\cos \left(x - \theta\right)$, & get,

$7 \cos x + 24 \sin x = R \cos x \cos \theta + R \sin x \sin \theta$.

$\text{Comparing the respective co-efficients of "cosx and sinx,}$

we have, $R \cos \theta = 7 \mathmr{and} R \sin \theta = 24$.

Squaring & adding , ${R}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = {7}^{2} + {24}^{2} ,$

$\mathmr{and} , {R}^{2} = {25}^{2} , \text{ giving, } R = + 25. . . \left[\because , R > 0\right]$.

Now, R=25rArrcostheta=7/R=7/25, &," similarly, "sintheta=24/25.

Alternatively, $\tan \theta = \frac{24}{7.} \therefore \theta = \arctan \left(\frac{24}{7}\right)$.

Altogether, we have,

$7 \cos x + 24 \sin x = 25 \cos \left(x - \theta\right) , \theta = \arctan \left(\frac{24}{7}\right)$.

Knowing that, $- 1 \le \cos \left(x - \theta\right) \le 1 ,$ we, on multiplication

by $25 > 0$, have,

$- 25 \le 25 \cos \left(x - \theta\right) \le 25 ,$

$i , e . , - 25 \le 7 \cos x + 24 \sin x \le 25$.

Adding $- 12 , - 37 \le 7 \cos x + 24 \sin x - 12 \le 13$.

Clearly, $\min . \left\{7 \cos x + 24 \sin x - 12\right\} = - 37 , \mathmr{and} ,$

$\max . \left\{7 \cos x + 24 \sin x - 12\right\} = 13$.