Difficult trigonometry problem to find the minimum value Find the minimum value of $5\cos A + 12\sin A + 12$.
I don't  know how to approach this problem. I need help.
I'll show you how much I got...
$$5\cos A +12\sin A + 12 = 13(5/13\cos A +12/13\sin A) + 12$$
 A: $5,12$ and $13$ are side of right triangle.
if $\sin\alpha=\frac{5}{13}$, then $\cos\alpha=\frac{12}{13}$.
We have :
$$13(\frac{5}{12}\cos A+\frac{12}{13}\sin A)=13(\sin\alpha\cos A+\cos\alpha\sin A)=13\sin(\alpha+A)$$
A: Denote $\varphi$ a solution of the system $\;\begin{cases}\cos\varphi=\frac{12}{13},\\\sin\varphi=\frac{5}{13}.\end{cases}\;$ Rewriting the function as 
$$13\sin(A+\varphi)+12$$
shows its bounds are $12-13=-1$ and $12+13=25$.
A: $(p\cos A+q\sin A)^2+(p\sin A-q\cos A)^2=p^2+q^2$  See Brahmagupta-Fibonacci Identity
$\implies(p\cos A+q\sin A)^2\le p^2+q^2$ the equality occurs if $p\sin A-q\cos A=0$
$\iff-\sqrt{p^2+q^2}\le p\cos A+q\sin A\le \sqrt{p^2+q^2}$
A: By C-S $$169=(5^2+13^2)(\cos^2A+\sin^2A)\geq(5\cos{A}+12\sin{A})^2,$$
which gives
$$-13\leq5\cos{A}+12\sin{A}\leq13$$ and 
$$5\cos{A}+12\sin{A}+12\geq-1.$$
The equality occurs for $(\cos A,\sin A)||(5,12).$
Id est, the answer is $-1$.
A: $5\cos A+12\sin A$ is the dot product of the vectors $(5,12)$ and $(\cos A,\sin A)$, of respective lengths $\sqrt{5^2+12^2}$ and $1$. The given expression achieves the extreme values $12\pm13$ when they are parallel and antiparallel.
A: As we know that,
$−\sqrt{a^2+b^2} \le a \cosθ + b \sinθ \le \sqrt{a^2+b^2}$
So for $5 \cos θ+12 \sin θ$ we have,
$−\sqrt{5^2+12^2} \le 5cosθ+12sinθ \le \sqrt{5^2 +12^2}$ 
= $−13 ≤ 5cosθ +12sinθ ≤ 13$
Now adding 12 we get, 
= $−13 + 12 ≤ 5 \cosθ + 12 \sinθ + 12 ≤ 13 + 12$
= $− 1 ≤ 5 \cosθ + 12 \sinθ + 12 ≤ 25$
So minimum value = −1
And maximum value = 25
A: The classical way is by canceling the derivative,
$$-5\sin A+12\cos A=0$$ or $$\tan A=\frac{12}5.$$
Then
$$\cos A=\pm\frac1{\sqrt{1+\tan^2A}}=\pm\frac5{13},\sin A=\pm\frac{\tan A}{\sqrt{1+\tan^2A}}\pm\frac{12}{13}.$$
The minimum is
$$12-5\frac5{13}-12\frac{12}{13}=-1.$$
