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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... $$5\cos A +12\sin A + 12 = 13(5/13\cos A +12/13\sin A) + 12$$

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    $|a\cos x+b\sin x|\leqslant\sqrt{a^2+b^2}$. – 2017-01-31

7 Answers 7

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$5,12$ and $13$ are side of right triangle.

if $\sin\alpha=\frac{5}{13}$, then $\cos\alpha=\frac{12}{13}$.

We have :

$$13(\frac{5}{12}\cos A+\frac{12}{13}\sin A)=13(\sin\alpha\cos A+\cos\alpha\sin A)=13\sin(\alpha+A)$$

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    but how does that help in finding the minimum value? @MoNtiDeaD MoonDogs – 2017-01-31
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    I show $5\cos A+12\sin A = 13\sin(A+\alpha)$. So the minimum is $-13+12 = -1$ – 2017-01-31
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    for $A+\alpha =A+arctan(5/12)= - \pi /2 +2k\pi$ – 2017-01-31
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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 $12+13=25$.

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$(p\cos A+q\sin A)^2+(p\sin A-q\cos A)^2=p^2+q^2$ See Brahmagupta-Fibonacci Identity

$\implies(p\cos A+q\sin A)^2\le p^2+q^2$ the equality occurs if $p\sin A-q\cos A=0$

$\iff-\sqrt{p^2+q^2}\le p\cos A+q\sin A\le \sqrt{p^2+q^2}$

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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 $-1$.

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$5\cos A+12\sin A$ 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.

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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

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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

$$12-5\frac5{13}-12\frac{12}{13}=-1.$$