How to find maximum value of trig function?
How to find maximum value of this: $$y = 5\sin x - 12\cos x$$
And I am more intrested in solving process, rather than answer. I know the answer. I am familiar with derivatives, not so good, but as I found prime of this function: $$y' = 5\cos x+12\sin x$$ but I dont know what to do next.
P.S Excuse me if grammar was poor, not a native english speaker.
$\endgroup$ 14 Answers
$\begingroup$Setting $y^{\prime}=0$ gives $5\cos x+12\sin x=0$, so $12\sin x=-5\cos x$ and dividing by $12\cos x$ gives
$\tan x=-\frac{5}{12}$. Then $\sec^{2}x=1+\tan^{2}x=\frac{169}{144}$, so $\sec x=\pm\frac{13}{12}$ and $\cos x=\pm \frac{12}{13}$.
Then $\sin x=(\tan x)(\cos x)=\mp\frac{5}{13}$, so $y=\pm13$ at the critical points.
$\endgroup$ 0 $\begingroup$i will finish what copper.hat is suggesting. we rewrite $$5\sin x - 12 \cos x = 13\left(\frac5{13}\sin x - \frac{12}{13}\cos x\right) = 13\sin (x-b), b = \sin^{-1}\left(\frac{12}{13}\right) \tag 1$$
now you can read off the maximum of value of $5\sin x - 12 \cos x$ over any interval of length $2\pi$ is $13$ and it happens at $x = b +\pi/2.$
$\endgroup$ 2 $\begingroup$You can write $a \cos x + b \sin y$ as $c \cos z$ for appropriate $c,z$ This gives a periodic function which is straightforward to extremise.
We can write $5 \sin x-12 \cos x = 13 ({5 \over 13} \sin x - {12 \over 13} \cos x)$, so let $\phi$ be such that $\cos \phi = {5 \over 13}, \sin \phi = {12 \over 13}$ (such a $\phi$ exists since $5^2+12^2 = 13^2$). Then $5 \sin x-12 \cos x = 13 \sin(x-\phi)$.
Since the range of $\sin$ is $ [-1,1]$, we see the maximum value is 13.
We see that the maximum value is attained for $x$ of the form $ \phi + (2n+1){\pi \over 2}$.
We have $\phi = \arctan {12 \over 5}$.
$\endgroup$ 5 $\begingroup$Hint: This is equivalent to solving a system of two equations in two unknowns, $5a+12b=0$, and $a^2+b^2=1$.
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