Trigonometric Functions

5cosA +12sinA + 12 = 13(5/13 cosA +12/13sinA) + 12
Now, for any values of B we can get sinB = 5/13 and we can replace cosB = 12/13.
We see that our assumption is right because we satisfy the condition sin^2B + cos^2B = 1 so we get 13(sinBcosA +cosBsinA) + 12 =13(sin(A+B))+12.
Therefore we know that minimum value of sinx=-1 and greatest is 1. Тhe greatest value is when sin(A + B) = 1 then value of the expression becomes 13.1 + 12 = 25

It is enough to use the inequality $ac+bd \leq \sqrt{a^2+b^2} \cdot \sqrt{c^2+d^2}$ ( Cauchy-Schwarz inequality for $n=2$ ).

In this case $a \cdot \sin \alpha + b \cdot \cos \alpha \leq \sqrt{a^2+b^2} \cdot \sqrt{\sin^2 \alpha + \cos^2 \alpha}=\sqrt{a^2+b^2}$

Differentiating the function 5sinA +12cosA + 12 and equating it to 0 gives critical point at tanA=12/5. We can easily judge that it is the maxima by double derivative test. Thus for maximum value sinA=12/13 and cosA=5/13, which gives 5sinA + 12cosA + 12=25.