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this is a silly question but here I go.

I was solving a question that required evaluating the derivative of an equation, which would result in finding a local min and a local max. The two points are in Quadrant 2 and 3,respectively.

The original equation is $\frac{\cos(2x+3)}{2e^x}$.

It came down to $\tan(2x+3) = -1/2$. After this, $\arctan(-1/2) = -0.463647..$. This lies in Q4, but I was wondering if I could use its complimentary angle (i.e., $2π - 0.463657$). The answer would be same, since $\tan(2π - \theta) = - \tan(\theta)$.

However, both yields different $x$ values and thus different answers.

I would be appreciated if you could explain the problem in my reasoning to me.

Gary
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Gtexx
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2 Answers2

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  1. Here is the correct solution:

    $$\tan(2x+3) = -1/2\\ \iff 2x-3=n\pi+\arctan\left(-\frac12\right)\:\:\text{for some } n\in\mathbb Z\\ \iff x=\frac{n\pi-3-\arctan\frac12}2\:\:\text{for some } n\in\mathbb Z$$

    finding a local min and a local max. The two required points are in Quadrant 2 and 3, respectively.

    Since the original function is differentiable on $\mathbb R,$ the required points are necessarily some of the above stationary points. And since $\tan$ is monotonic around its roots, the above stationary points alternate between minimum and maximum points.

    Plugging in $n=-1$ and $n=0$ gives $x=-3.30$ and $x=-1.73,$ which are a quadrant-2 minimum and a quadrant-3 maximum, respectively.

    Hence, the complete solution set is $$\left\{\frac{n\pi-3-\arctan\frac12}2\,\Bigg\vert\,n\equiv-1,0\pmod4\right\}.$$

  2. It came down to $\tan(2x+3) = -1/2$. After this, $\arctan(-1/2) = -0.463647..$. This lies in Q4,

    Yes, by definition and convention, $\arctanα$ lies in the interval $\left(-\frac\pi2,\frac\pi2\right),$ thus $$\tan\alpha=p\kern.6em\not\kern-.6em\implies\alpha=\arctan p;$$ for example, try $\alpha=\pi$ which is in neither quadrant $4$ nor $1.$

    but I was wondering if I could use its complimentary angle (i.e., $2π - 0.463657$). The answer would be same, since $\tan(2π - \theta) = - \tan(\theta)$.

    The complementary angle of $θ$ is $\displaystyle\left(\frac\pi2-θ\right)$ rather than $\left(2\pi+θ\right).$

    And yes, $$\tan\alpha=p\kern.6em\not\kern-.6em\implies\alpha=2\pi+\arctan p,$$ as you've just illustrated.

    However, both yields different $x$ values and thus different answers.

    Yes, your two suggested answers correspond to $n=0$ and $n=2,$ which give the points $x=-1.73$ and $x=1.41,$ respectively. Only the former (out of infinitely many points) satisfies this exercise's requirement.

ryang
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  • Hello, thanks for your answer. You plugged in 'n = -1 and n = 0' to arrive at the correct answers. But, how can one determine which 'n' to plug in to attain correct answers? – Gtexx Feb 07 '22 at 08:15
  • @Gtexx Thanks for the observation; I've edited the answer. P.S. Do click on the link. – ryang Feb 07 '22 at 09:04
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Note the tangent function has a period of $\pi$. Hence,

$$\begin{align}\tan(2x+3)&=-1/2\\ \implies 2x+3&=\arctan(-1/2)+n\pi,\quad n\in\mathbb Z\\ &\approx -0.46365+n\pi,\quad n\in\mathbb Z\\ \implies x&\approx -1.73+n\frac{\pi}{2},\quad n\in\mathbb Z\end{align}.$$

Note $x$ is in quadrant I,II,III, IV when $n\equiv 2,3,0,1 \quad (\mod 4), $ respectively.

Golden_Ratio
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