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$$-12\sinθ-5\cosθ=0$$

Is it possible? or do I need to generate a graph? I heard the Newton Raphson Method can be used...not sure how to proceed please advise.

I'm trying to calculate the minimum and maximum values of: $y=12\cosθ-5\sinθ$

The equation above is a derivitive of $y$.

Modrisco
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5 Answers5

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Hint: Check if $\theta=\pi/2,3\pi/2$ (I'm considering the interval $[0,2\pi]$ ) are solutions, and then divide by $\cos$ and solve for $\tan$.

Vincenzo Oliva
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$$12\sin\theta+5\cos\theta=0\\\implies\frac{12}{13}\sin\theta+\frac5{13}\cos\theta=0$$ Now let $\cos a=\frac{12}{13},\sin a=\frac{5}{13}$. This works because $\sin^2a+\cos^2a=1$. We now have $$\cos a\sin\theta+\sin a\cos\theta=0\\\implies\sin(\theta+a)=0 \\\implies\theta=\pi n-a$$ for integer $n$.

Pauly B
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Yes, it is possible, though the answer might not look nice. First divide by $-12$: $$\begin{align}&\sin\theta + \frac 5{12}\cos\theta = 0\implies \\ \implies &\sin\theta = -\frac 5{12}\cos\theta \\[2ex] \implies &\sin^2\theta = \frac{25}{144}\cos^2\theta \\[2ex] \implies &1 - \cos^2\theta = \frac{25}{144}\cos^2\theta\end{align}$$ Can you take it from here?

GPerez
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I don't think you should differentiate.

Your $y$ can be written as $$ y(\theta)=13 \cos(\theta+\arctan(5/12)). $$ Can you see from this what the maximum and minimum values are?

mickep
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  • I didn't do it that way, but something like it, I calculated the arctan of -5/12...then I minused that value from 180 and 360 degrees to get the min and max points...then subed those values back into y, and received the answers -13 and 13....I'm mind boggled here but I got the answer at the back of the book anyway! – Modrisco Jun 16 '15 at 12:56
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y = 12cosx -5sinx

y'= -12sinx -5cosx

put y'=0

-12sinx -5cosx = 0

12sinx + 5cosx =0 (multiplied by -1 both sides )

12sinx = -5cosx

sinx/cosx=-5/12

tanx=-5/12

X = tan inverse (-5/12)

Now can you proceed?

Shankar Subramanian
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