This is Simple....

Maximum value of trigonometric function of the form $a\sin\theta + b\cos\theta + c$ is always equal to $\sqrt{a^{2}+b^{2}}+c$ .

Putting the values of a,b,c with their respective sign, we get the maximum value of this function, which is equal to $\boxed{-5}$ .

by cauchy schwarz inequality ... (5^2 + 12^2)(sin^2 + cos^2) => (5sin + 12cos)^2

(5sin + 12cos) <= 13

13 - 18 = -5

a sin(x) + b cos(x) = R sin (x+y), problem solved.

Let f(t) = 5sin(t) + 12cos(t) - 18. Then f'(t) = 5cos(t) - 12sin(t) = 0, or tan(t) = 5/12. So sin(t) = 5/13, cos(t) = 12/13, and 5sin(t) + 12cos(t) - 18 = 25/13 + 144/13 -18 = 169/13 - 18 = 13 - 18 = -5. Ed Gray

For the value of function to be maximum

$\frac { d }{ d\theta } (5sin\theta +12cos\theta -18)=0$

or $5cos\theta =12sin\theta$

or $tan\theta =\frac { 5 }{ 12 }$ or $\theta \approx 22.61$

Putting this value in our function we get

$5sin(22.61)+12cos(22.61)-18\approx -5$

Let x = sin(theta), then the equation can be turned into algebra where you have 5x + 12 ( 1- x^2)^(1/2) - 18 .

Taking a derivative and solving quickly gives:

5( 5/13) + 12( 12/13) -18 = -5